International Mathematics Olympiad
WORK BOOK INSTANT
4
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Chapter-1 : NUMBER SENSE 1.
(D) : 100 tens = 100 × 10 = 1000 = 1 thousand
2.
(B) : (A) 10 tens = 10 × 10 = 100 (B) 100 tens = 100 × 10 = 1000 (C) 1 hundred = 100 (D) 100 ones = 100 So, 100 tens is different from others.
18. (A) : 73,842 has 30 hundreds as 30 hundreds = 30 × 100 = 3000. 19. (A) : 195 is rounded off to 200. 20. (B) : Descending order of numbers means arranging the numbers starting from greatest and proceeding towards least. The order of digits 5, 2, 4, 1, 3 in descending order is 54321.
3.
(B) : 1 million = 10 lakh.
4.
(C) : 1 crore = 100 lakhs = 10 million.
5.
(B) : Fifty six thousand four hundred fifty is same as 56,450.
6.
(A) : Smallest five digit number using 9, 0, 5, 2, 4 digits only once is 20459.
7.
(B) : Largest five digit number is 95420.
8.
(A) : 24590 is the smallest five digit number having 0 at one’s place.
9.
(D) : Largest five digit number having 0 at thousand’s place is 90542.
10. (C) : Place value of 8 in the number 78045 = 8000 Place value of 0 in the number 78045 = – 000 Difference between the place values = 8000 11. (B) : Expanded form of 52946 is 5 × 10000 + 2 × 1000 + 9 × 100 + 4 × 10 + 6 × 1 = 50000 + 2000 + 900 + 40 + 6. 12. (B) : The ascending order of numbers means arranging the numbers starting from the least and proceeding towards the greatest. So, ascending order of 9411, 9145, 9015 and 9149 is 9015, 9145, 9149, 9411. 13. (D) : Largest five digit number having 0 at ten’s place and without repeating the digits is 98706. 14. (B) : 7 thousand 9 hundred is 7 × 1000 + 9 × 100 = 7000 + 900 = 7900 15. (A) : Place value of 8 in 76,863 is 800. 16. (B) : 50908 = 50 thousands + 90 tens + 8 ones. 17. (B) : Of the given numbers 23189, 32189, 31218 and 31289 the number 32189 is the greatest number.
21. (C) : A is the smallest of all the other numbers because it has the smallest digit as compared to others at ten’s place. 22. (C) : (A) X V = 10 + 5 = 15 (B) XI X = 10 + 9 = 19 (C) I X IV = Meaningless (D) X X IV = 10 + 10 + 4 = 24. 23. (C) : X can be subtracted from L, M and C only. 24. (D) : CCC = 100 + 100 + 100 = 300. 25. (B) : CM = 1000 – 100. 26. (D) : LII = 50 + 2 = 52. 27. (A) : M > 100. 28. (A) : The All Sports magazine sold 43,810 copies and it has 4 in ten thousand’s place. 29. (A) : 34519 lies between 34500 and 34600. It is near to 34500, so it is rounded off as 34500 to the nearest hundred. 30. (B) : 7952, 5642, 3241, 2341 are arranged in descending order i.e. numbers in order from greatest to least. 31. (B) : ` Ninety four thousand = ` 94,000 32. (D) : 5,559 has 9 in the one’s place. 33. (A) : 81521 lies between 81500 and 81600. It is nearer to 81500 so it is rounded off as 81500 to the nearest hundred. 34. (A) : 73,862 has 7 in ten thousand’s place. 35. (C) : The largest four digit number using the digits 7, 9, 4, 5, if Tina put 9 in the ten’s place = 7594.
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Class 4
3
Chapter-2 : COMPUTATION OPERATIONS 1.
2.
(A) : 12 × 89 = 89 × 12 is true as multiplication of two numbers is commutative. (A) : So,
= 4 hamburgers = 8 hamburgers
= 3 cupcakes = 4 × 3 = 12 cupcakes Number of cupcakes more than hamburgers = 12 – 8 = 4. 3.
(A) :
4.
(A) : Number of pencils Sunil has = 383 Number of pencils he gave to his friend = 9 Number of pencils left with him = 383 – 9 = 374
5.
12. (C) : Total number of seats = 145 Number of rows = 5 Number of seats in 1 row = 145 ¸ 5 = 29 13. (D) : Second multiple of 9 is 18 Now, 18 × 12 = 216. 14. (A) : S – 100 = T – 100 S – T = – 100 + 100 Þ S – T = 0 Þ S = T
(B) : Number of people at the fair on Saturday = 3136 Number of people at the fair on Sunday = +3759 Total = 6895 6895 is rounded off to 7,000. (B) : 7 × 3 = 21
7.
(C) : Since Samrat has 6 yards long string. So, first he will convert yards to feet. 1 yard = 3 feet So, 6 yards = 6 × 3 = 18 feet
8.
(B) : 65 ´ 100 = 6500
9.
(B) : Number of people living in Delhi in 1990 = 455268 Number of people living in Delhi in 2000 = 426599 Number of people decreased in 10 years = 455268 – 426599 = 28669
10. (C) : Factors of 12 = 1, 2, 3, 4, 6, 12 Factors of 15 =1, 3, 5, 15 Common factors of 12 and 15 = 1 and 3. 11. (A) : Number of cans in 1 box = 24
= 3 Þ 3
15. (B) :
= 7 Þ Þ
= 3 Þ
= 1
+ 1 + 1 = 7
= 7 – 2 = 5 = 20 and
shows 3 × 2
6.
4
Number of cans in 8 boxes = 24 × 8 = 192
= 100
So, the value of +
+
+ = 100 + 100 + 100 + 100 + 20 + 20 +5 + 5 + 5 + 1 + 1 + 1 + 1 = 459. 16. (C) : Distance covered on Thursday =167 miles Distance covered on Friday = 68 miles Distance covered on Saturday = 73 miles \ Total distance covered = 167 + 68 + 73 = 308 miles Approximately he drove 308 » 300 miles on three days. 17. (B) : Number of pages Jyoti wants to read = 650 Number of pages Jyoti has already read = 541 Number of pages to be read more = 650 – 541 = 109 18. (A) : Divisor = Quotient × Dividend + Remainder 1364 = (454 × 3) + 2 19. (C) : Number of seats in 1 truck = 1025 Number of seats in 41 trucks = 1025 × 41 = 42025 20. (A) : Number of pages read by Monika in 7 days = 210 Number of pages read in 1 day = 210 ¸ 7 = 30 pages IMO Work Book Solutions
21. (B) : As we know,
Þ
= 5 The value of
= `140
–
or Train = `140 + Car Now, we are given
.... (i)
+
+
+
= `1280 (from (i))
2 Cars + 2(140 + Car) = `1280 or 4 Cars + 280 = `1280 4 Cars = `1000 1000 = `250 Þ 1 Car = ` 4 Now, we also know, +
.... (ii)
= `820
+
or Car + (140 + Car) + Bear = `820 .... [from (i)] 2 Cars + Bear + 140 = 820 or 2 × 250 + Bear + 140 = 820 .... [from (ii)] \ Bear = 820 – 500 – 140 = 180 Bear = `180 Now, required cost of +
+
25. (B) : Number of balls in jar A = 121 Number of balls in jar B = 37 Number of balls in jar C = 180 Total number of balls in three jars = 121 + 37 + 180 = 338 So, correct number is 340. (Closest to the number of balls in the jars) 26. (B) : Number of racks of books = 215 Number of books in 1 rack = 303 Number of books in 215 racks = 303 × 215 = 65145 27. (D) : Cost of 18 tickets = ` 86,886 86886 = ` 4,827 Cost of 1 ticket = 18 28. (B) : 18 3
6 3
2
29. (B) : Smallest factor of a number is 1. 30. (A) : The number which has 0 or 5 in one’s place is divisible by 5. So, 5105 will be divisible by 5.
+
= 2 × 250 + 2 × 180 = ` 860 22. (C) : Total number of balls Shiv bought = 7 × (5 + 3)
31. (C) : Factors of 18 = 1 , 2, 3 , 6, 9 , 18 Factors of 27 = 1 , 3 , 9 , 27 6 is not a common factor of 18 and 27.
23. (C) :
32. (B) : 8, 16, 24, 32, 40, 48 are all multiples of 8.
Þ
+
+
+
= 1240
4
= 1240 1240 Þ = = 310 4 Now, 1000 – =
33. (A) : The numbers in section (ii) shows that the number can be divided by 3 and 5. 34. (B) : The numbers that can be divided by both 2 and 3 can also be divided by 6.
1000 – 310 =
Þ = 690 So, the value of 24. (D) : 5
35. (D) : 8 –
Also,
+ + + = 125 125 = = 25 5 – 4 =
Þ
= 4
Þ
25 = 5
Þ
= 690 – 310 = 380
+
Þ
= 40 = 5
= 125 Also, Þ
5 + 3
Þ
3
+
+
+
= 155
= 155
+ = 150 Þ
= 50.
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Class 4
5
Chapter-3 : FRACTIONS 1.
(B) : Number of equal parts = 4 Shaded part = 1
12. (C) : * -
1 4 Number of equal parts = 4 Shaded parts = 3 3 Required fraction = 4 Fraction of shaded parts of both the figures 1 3 = + 4 4 Required fraction =
2.
3. 4. 5.
6. 7. 8.
9.
(B) : The numerator of a fraction is 2 more than 6 its denominator. So, fraction is . 4 æ 1 + 1 = 1 ö (A) : 2 halves makes a whole. è 2 2 ø (B) :
19 7 12 = 20 20 20
23 is a fraction where numerator (23) is 22 greater than denominator (22).
16. (A) :
3 5 7 9 11 , , , , are in ascending order. . 6 6 6 6 6
`\
2 ´ 4 8 1´ 3 3 = = and 3 ´ 4 12 4 ´ 3 12 8 3 > 12 12 2 1 > . 3 4 1 5 kg = kg 2 2
1 kg 4 1 5 Weight of peas = 1 kg = kg 4 4 5 1 5 Total weight = + + 2 4 4 10 + 1 + 5 16 = = = 4 kg 4 4
Weight of tomatoes =
2 6 2 7 1 18 + + + + = 11 11 11 11 11 11 So, other arm is (B) : Sum of
18 6 2 5 3 = +x+ + + 11 11 11 11 11 18 16 18 16 2 = + x Þ x = = . 11 11 11 11 11
6
5 1 = 10 2 Half of the fish in aquarium are gold fish.
18. (B) : Weight of potatoes = 2
1 ´ 5 = 1 5 So, 5 onefifths will make one whole.
20 4 = 25 5
1 1 2 + = 4 4 4
15. (A) : Fraction of gold fish =
Þ
(A) :
11. (A) :
=
14. (A) : Number of organisms in aquarium = 4 + 5 + 1 = 10 Number of gold fish = 5 5 \ Required fraction = 10
Now,
1 46 (B) : 5 = 9 9 2´3 6 (A) : 5 ´ 3 = 15
3 2 1 - = which is equal to 5 5 5
+
17. (A) : Converting the fractions into like fractions.
(A) :
10. (C) :
13. (B) :
7 5 5 7 12 = Þ *= + = 19 19 19 19 19
19. (B) : 3 times of .
25 25 75 25 = 3 ´ = = 81 81 81 27
5 2 10 5 ´ = = 4 3 12 6 21. (D) : Total number of equal parts = 10 Unshaded parts = 4 20. (C) :
IMO Work Book Solutions
\
Required fraction =
4 10
29. (D) : Fraction of red fabric =
1 m 6 Fraction of red fabric bought more than green 5 1 4 fabric = - = m. 6 6 6
22. (C) : Total number of equal parts = 32 Unshaded parts = 16 16 1 = \ Required fraction = 32 2
Fraction of green fabric =
23. (A) : Total number of equal parts = 10 Shaded parts = 5 5 \ Required fraction = 10 24. (D) :
30. (A) 3 31. (A) : Hanit won exactly of the ribbons for the 5 I st place means Hanit won 3 ribbons for I st place out of 5 ribbons.
2 2 < 4 3
25. (B) : Number of cards in the bag = 12 Number of horse cards = 4 4 \ Required fraction = 12
32. (B)
3 3 = 2 4 4 27. (A) : Total number of marbles in the bag = 8 + 4 + 5 + 9 = 26 Number of red marbles = 8 8 \ Required fraction = 26
26. (C) : The model shows 1 + 1 +
28. (D) : Number of green marbles = 5 5 \ Required fraction = 26
5 m 6
33. (B) : Total number of students = 8 Number of students whose name starts with ‘A’ = 3 3 \ Required fraction = 8 34. (B) : Total number of alphabets in word C E L E S T I N E made from tiles = 9 Number of vowels = 4 4 Required fraction = 9 35. (D) : Total number of equal parts = 12 Unshaded parts = 5 5 Required fraction = 12
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Class 4
7
Chapter-4 : LENGTH, WEIGHT, CAPACITY, TIME & MONEY 1.
(D) : Time taken by Amrita to reach school is 35 mins.
2.
(B) : A year will be a leap year if it is divisible by 4 but not by 100. If a year is divisible by both 4 and 100, it is not a leap year unless it is also divisible by 400. 2000 2000 = 500 and = 20 4 100 2000 = 5 So, 400
8.
(A) : Distance travelled by car = 75 km 800 m = 75800 m Distance travelled by air = 1800 km 40 m = 1800040 m Total distance travelled in all =75800 + 1800040 =1875840 m So, total distance travelled =1875 km 840 m.
9.
(B) :
Now,
= `2 Þ
+
= `1
...(i)
Hence, 2000 is a leap year. 3.
(D) : 7 years 2 months = 7 × 12 + 2 = 86 months. (As we know, 1 year = 12 months).
4.
(A) : As we know, = 19 Þ
+ 5
+
+ 3
= 19 –
...(i)
= 94
...(ii)
+ = 22 Þ = 22 – ...(iii) Substituting the values of (i) and (iii) in (ii), we get Þ
5
+ 22 –
Þ
4
+ 22 + 57 – 3
Þ Þ
+ 3(19 –
) = 94
= 94
+ 79 = 94
+
+
+
+
= `9
...(iii)
= `2 Now,
+
So,
+
= `9
= 9 – 4 = `5
= `9
`1 + `5 + in (i) we get
(C) : Months having only 30 days = April, June, September and November. So number of months = 4.
6.
(C) : 1000 times of 1 metre = 1 kilometre. (Since, 1 km = 1000 m)
7.
(A) : 38 km 43 m = 38 × 1000 + 43 = 38043 25 km 3 m = 25 × 1000 + 3 = 25003 \ 38 km 43 m + 25 km 3m = 38043 + 25003 = 63046 m or 63 × 1000 + 46 = 63 km 46 m.
8
...(ii)
From (ii) and (iii), we get
= `3.
= 19 – 15 = 4 kg. 5.
= `11 1
Substituting these values in (iii), we get
=15
Putting the value of
+
10. (A) : As we know, –
= 20 g or
=
– 20 .... (i)
and –
= 70 g or
=
+ 70 ...(ii)
10
+ 10
+ 10
= 3200 g
IMO Work Book Solutions
10 [
+ 70] + 10
+ 10 (
30
+ 700 – 200 = 3200
30
= 3200 – 500 = 2700
– 20) = 3200
= 90 From (i), we have = 90 – 20 = 70 g. 11. (B) : Cost of 500 g apples = ` 50 Cost of 1000 g (1 kg) apples = 50 × 2 = ` 100. 12. (B) : 2 kg = 2000 grams (As we know, 1 kg = 1000 g) So, number of 500 sugar packets required to 2000 = 4. make 2 kg = 500 13. (B) : Cost of 1000 g grapes = ` 80 80 Cost of 1 g grapes = 1000 80 ´ 250 = ` 20. Cost of 250 g grapes = 1000 14. (A) : Weight of 6 balls = 1500 g 1500 Weight of 1 ball = = 250 g. 6 15. (B) : 1000 mL = 250 mL + 250 mL + 250 mL + 250 mL. 16. (B) : Weight of 9
= 18 g
18 ...(i) = 2 g 9 Weight of an apple + 3 = 28 g Weight of an apple + (3 × 2)g = 28 g or, Weight of an apple + 6 g = 28 g \ Weight of an apple = (28 – 6)g = 22 g
Weight of 1
22. (B) : Cost of 4000 g of potato = `36 36 Cost of 1 g of potato = ` 4000 Cost of 500 g of potato = 36 ´ 500 = ` 4.50. 4000 23. (B) : Cost of a shirt = `120 Cost of a pant = `150 Cost of a tie = `50 Total cost of 3 items = `320 Total amount of money Mr. Verma has = `500 \ Amount of money Mr. Verma gets back = `(500 – 320) = `180. 24. (C) : Cost of 1 soup = `30 Cost of 2 soups = `(30 × 2) = `60 25. (C) : Cost of 2 packets of chips = `(10 × 2) = `20 Cost of 1 pizza = `70 Cost of 2 juices = `(10 × 2)= `20 \Total cost Rohan has to pay = `(20 + 70 + 20) = `110. 26. (A) : Cost of 5 cups of tea = `(2 × 5) = `10 Cost of 1 pizza = `70 Total cost Rohit should have = `(10 + 70) = `80 +
27. (D) :
3
= 54
Þ
=
Now,
+
=
17. (B) : Capacity of mug = 3 litres 200 mL = 3200 mL Capacity of cup =750 mL \ Capacity that mug can hold more than cup = (3200 – 750)mL = 2450 mL.
+
= 54
54 = ` 18 3 +
Þ
+ 18 + 18 = 196
Þ
= 196 – 36 = 160
= ` 196
18. (C) : Weight of P is 1700 g. 19. (C) : The weight of Q is 2600 g.
Also,
+
+
= 240
20. (C) : P is lighter than Q by (2600 – 1700) = 900 g. 21. (B) : Total weight of P and Q is 1700 + 2600 = 4300 g. Class 4
Þ
+ 18 + 160 = 240
9
Þ
or 3 Necklaces = `1560 1 Necklace = `520 Since, 1 Bangle = `1290 – 2 Necklaces \ 1 Bangle = `(1290 – 1040) = `250 As we know,
= 240 – 178 = ` 62
Finally,
+
+
= ` 267 + \
Þ
+ 160 + 62 = 267
\
28. (B) : 1 kg = 1000 g 1 1000 kg = = 500 g. Þ 2 2 29. (A) : 2 nd Tuesday falls on 8 th August So, a day before 8 th August is 7 th August.
+
= `1290
+
= `550
= `180
32. (B) : Weight of watermelon = 11 kg Weight of book = 1 kg Weight of sugar bag = 5 kg Weight of box = 7 kg So, the lightest item is the book. 33. (D) : Sonika left the house at 1 : 15 p.m. Time after 1 hr 45 mins will be 3 : 00 p.m.
or 1 Bangle = `1290 – 2 Necklaces .... (i) Now,
+
31. (D)
30. (B) : As it is given, +
+
or 1 Ring = `550 – (1 Bangle + 1 Earing) = `550 – 370 = `180
= 267 – 222 = ` 45.
+
= `890
1 Earing = `890 – (1 Necklace + 1 Bangle) = `890 – 770 Þ `120 Now,
Þ
+
34. (A) : 1 block = 2 balls 2 blocks = 4 balls = 10 cans.
= `1020
or 2( `1290 – 2 Necklaces) + 1 Necklace = `1020 (from (i)) Þ 2580 – 4 Necklaces + 1 Necklace = `1020
35. (C) : 4 th Thursday = 22 nd November 6 days after 22 nd November = 28 th November i.e., Wednesday.
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10
IMO Work Book Solutions
Chapter-5 : GEOMETRY 1.
2.
(B) : Number of line segments required to
16. (A)
B D (C) : The triangles formed are, A, B, C, D, E, F, G, H, BG, BE, CE, CG, i.e., 12 in number.
17. (D) : In IJKL, we have Þ IJ = LM + MK Þ 8 = 5 + MK \ MK = 3 cm In MNOP, we have MP = NO (as we know, in rectangle opposite sides are equal.) Þ MK + KP = NO Þ 3 + KP = 12 Þ KP = 9 cm.
A B H G
E F C D
3.
(D)
4.
(D)
5.
(B) : Cube have 6 faces and all of which are squares of same size.
6.
(C)
7. 8.
(A) (B) : Number of squares are 12, 34, 56, 78, 2367, 12345678 = 6 squares.
9.
Hence, Fig. II has the least perimeter.
A C E make is AB, BC, CD and DE = 4.
18. (D) 1
2 3
4
6 7 5
8
(D) : Perimeter of the given figure is 36 cm. 2 2 2 2 2
2 2 2 2 2
2 2 2 2 cm 2 2 2
2 10. (C) : Breadth = 6 cm Length = 10 + 6 = 16 cm Perimeter = 2(l + b) = 2(16 + 6) = 2 × 22 = 44 cm.
11. (A) : We have, AB + BC + CD = FE Þ 3 + 3 + CD = 12 Þ CD = 6 cm.
19. (B) : Kamal should know about the perimeter of the park to make sure he buys enough tiles. 20. (D) : Perimeter of the figure = 5 + 2 + 2 + 3 + 5 + 4 + 2 + 1 = 24 cm. 21. (A) : Perimeter of rectangle B = 2(length + breadth) = 2(8 + 4) = 24 cm Perimeter of square A = 4 × side Þ 24 = 4 × side Þ Side of square = 6 cm. 22. (C) : Points G and E appear to lie on the same line that is similar to line m. 23. (B) : Three triangles will be formed namely DGDE, DGEF and DGDF. G
12. (D) 13. (A) 14. (C) : Second odd prime = 5 Diameter of circle = 2 × radius = (2 × 5) units = 10 units. 15. (B) : Perimeter of a figure is the sum of its sides. Perimeter of Fig. I = 2 + 5 + 2 + 5 = 14 cm Perimeter of Fig. II = 2 + 2 + 2 = 6 cm Perimeter of Fig. III = 4 + 4 + 4 + 4 = 16 cm Class 4
D E F
24. (C) : Region 2 is in triangle and rectangle. Region 6 is in circle and rectangle. Region 5 is in triangle and circle. So, three numbers are exactly in the two regions. 25. (A) 26. (A)
11
27. (D) : Perimeter = 18 cm.
32. (D)
28. (D)
33. (C) : The lines that appear to be nonintersecting are in figure 1 and 2.
29. (C) : Varun made a cube.
34. (B) : Figure 2 and 3 show pairs of intersecting lines.
30. (A) 31. (C) : 18 cubes are required to make the given solid.
35. (C) : Figure 3 shows the intersecting angle is 90°.
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12
IMO Work Book Solutions
Chapter-6 : DATA HANDLING 1.
(A) : Gautham has `24 and Anusha has `48 So, the amount Anusha have to give = `[(48 – 24) ¸ 2]
10. (A) : Aditya earn in April = `4000
= `(24 ¸ 2)
Aditya earn in March = `1500
= `12. 2.
\ Required earning = `(4000 – 1500) = `2500
(B) : Total savings in the week = `[32 + 48 + 24 + 20 + 56]
5.
Now, Harsh saved money less than remaining four children by (160 – 20) = `140. (C) : In the month of February, 24 printers sold which is highest.
2000 + Feb + 1500 + 4000 + 4500 = `15000
Þ
Earning in Feb = `(15000 – 12000) = `3000
12. (A) : Amount earned in March = `1500 Thrice of it = `(3 × 1500) = `4500. From the bar graph we see that,
and number of printers sold in May = 12
Aditya earn `4500 in the month of May.
Collection of money in April = `(20 x 250) = `5000
and collection of money in May = `(12 x 250) = `3000 \
Required Amount = `(5000 – 3000) = `2000
(A) : Total sale of printers = 18 + 24 + 16 + 20 + 12 = 90.
7.
Þ
(A) : No. of Printers sold in April = 20
\
6.
Earning in (Jan + Feb + Mar + April + May) = `15000
(A) : Total savings except Harsh = `(180 – 20) = `160
4.
11. (C) : Total earning = `15000 Þ
= `180. 3.
Acc. to question, 20 + Another game = 42 \ Another game = 42 – 20 = 22 people. Now, we see that 22 people like Badminton.
(B) : No. of people like swimming = 20 No of people like basketball = 17 \ No of people like swimming more than basketball = (20 – 17) = 3.
13. (C) : Amount earned in May = 4500 æ 1 ö Amount saved = ` çè ´ 4500 ÷ø = ` 900 5 Amount of money spend = `(4500 – 900) = `3600. 14. (A) : Amount earned in April = 4000 1 ö æ Amount spent = ` çè 4000 ´ ÷ø = `800 5 \ Amount saved = `(4000 – 800) = `3200. 15. (A) No. of boys joined yoga in 2004 = 160 No. of boys joined yoga in 2005 = 200 \
Required no. of boys = 200 – 160 = 40.
16. (B) : No. of women joined yoga in 2004 = 480
8.
(C) : Cycling.
No. of women joined yoga in 2005 = 640
9.
(D) : No. of people like swimming = 20
\
Class 4
Required no. of women = 640 – 480 = 160.
13
26. (C) : No. of detective books he read = 4 × 5 = 20.
17. (D) : No. of girls joined in 2004 = 240 No. of women joined in 2004 = 480 \
27. (A) : No. of adventure books he read = 4 × 4 = 16.
Difference = 480 – 240 = 240.
18. (A) : No. of boys joined in 2005 = 200 \ Total membership fees of boys = `(19 × 200) = `3800 Also, no. of girls in 2005 = 180 Total fees of girls = `(8 ×180) = `1440 \ Total amount = `(1440 + 3800) = `5240
28. (A) 29. (C) : 4 students get up at 6 : 30 am. 30. (B) : No. of students wake up between 7 : 00 am to 8 : 00 am = 8 + 18 + 10 = 36. 31. (B)
19. (A) : In Mathematics, he got 100 marks.
32. (C) : 1999, 2000
20. (A)
33. (A) : Rainfall in 1997 = (3 × 20) cms = 60 cms
21. (A) : In Hindi, he got 30 marks.
Rainfall in 1996 = (8 × 20) cms = 160 cms
22. (A) : Anshu saved = `100
\
City receive 100 cms, less rainfall in 1997.
34. (A) : Rainfall in 1998 = (9 × 20) cms = 180 cms
and Suhana saved = `150 \ Anshu saved `50 less than Suhana.
Rainfall in 1999 = (5 × 20) cms = 100 cms
23. (A) : Amount saved by Anjali = `150
\
Amount spends on toy = `48
Required rainfall = (180 – 100) cms = 80 cms.
35. (B) : No. of visitors in September = 240 + 360 = 600
Amount left = `(150 – 48) = `102.
Now, æ1 1 of 600 ö = 6 ´ 600 çè ÷ø 5 5
24. (A) : Rizwan saved = `350 Mohit saved = `200 and twice of Mohit’s savings = `(2 × 200) = `400 \ Rizwan needs to save another (400 – 350) = `50 25. (C): Total no. of books = 4 × [4 + 4 + 1 + 4 + 5] = 4 × 18 = 72
= 6 × 120 = 720. Now, from graph we can see that, No. of visitors in October = 120 + 180 = 300 No. of visitors in November = 300 + 420 = 720 No. of visitors in December = 400 + 500 = 900 \ Required answer is November.
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14
IMO Work Book Solutions
Chapter-7 : LOGICAL REASONING 1. 2.
(D) : An arrow and line gets added in every alternate step.
12. (B) :
(C) : A = 1, B = 2, ................, Z = 26. 13. (A) : 10, 8, 6, 4, 2, 0
Now, E = 5
–2 –2 –2 –2 –2
PEN = 16 + 5 + 14 = 35
3.
4.
\
P = 16, A = 1, G = 7, E = 5
\
PAGE = 16 + 1 + 7 + 5 = 29
+2
+2
14. (A) : 5, 8, 7, 10, 9, 12, 11, 14 +2
(C) : The arrow ( ) moves 90° anticlockwise in every step and ( ) moves 90º clockwise in every step. (A) : Figure B is formed when the shape in figure A gets reduced by one line segment and a shaded circle shifts the unshaded ones to move out. So, figure (S) will resemble as follows
+2
.
5.
(C) : The figure gets rotated to 90° anticlockwise direction.
6.
(A)
7.
(C) : Epigene, Epilogue, Episode, Epitaxy, Epitome
+2
+2
15. (A) 16. (A) : 10, 100, 1000, 10000, 100000 ×10
×10
×10
×10
17. (B) : 2, 4, 6, 8, 10, 12 +2 +2 +2 +2 +2
18. (C) : The pattern is repeated after every 5 th shape. So, 13 th shape is ' ' . 19. (C) : Figure 5
Figure 6
3, 5, 2, 1, 4 20. (C) : 281, 278, 275, 272, 269, 266, 263 8.
–3
(A) :
–3
×2 ×2 ×2
9.
(C) :
–3
–3
–3
×2
22. (B) : The arrangement follows the given
10. (D) : After interchaning, the new number will be 4 9 1 3 8 6 5 7 Left
–3
21. (A) : 5, 10, 20, 40, 80
pattern :
,
23. (B) :
Right
The seventh digit from left end is 5 and the third to the left of 5 is 3. 24. (A) 11. (D) :
Class 4
25. (A)
15
30. (D)
26. (D) : 46980, 49480, 51980, 54480 +2500
+2500
31. (C)
+2500
27. (A) : If we are thirsty we drink water and in other planet water is called light. So, if someone is thirsty there, he drink light.
32. (A)
28. (B) : 3, 6, 12, 24, 48, 96
34. (D)
×2 ×2 ×2
×2 ×2
29. (C) : The pattern is repeated after every 3 rd
33. (B) : 3 rd letter from right end is X, and 9 th from left of X is O. 35. (C) : If 17 th falls on Tuesday. Then, 16 th must be Monday and 30 th will be Monday, i.e., 5 th Monday.
shape. So, vvv
16
IMO Work Book Solutions
2012 - 6th SOF IMO 1. (B) : Because, Q has larger sides than that of others. 2. (A) 3. (D) : 8800,
Q P 10200, 11600 , 13000, 14400 , 15800
+1400
+1400
+1400
+1400
+1400
4. (B) : House A = 5000 + 700 + 25 × 1 = 5725 House B = 7000 + 72 × 10 + 115 × 1 = 7000 + 720 + 115 = 7835 House C = 50 × 10 + 99 × 1 = 599 House D = 40 × 100 + 35 × 10 + 5 × 1 = 4355 ∴ House B has the largest value 5. (A) : Let the twelve friends be A, B, C, D, E, F, G, H, I, J, K and L Now, Handshakes with A = AB, AC, AD, AE, AF, AG, AH, AI, AJ, AK, AL Handshakes with B = BC, BD, BE, BF, BG, BH, BI, BJ, BK, BL Handshakes with C = CD, CE, CF, CG, CH, CI, CJ, CK, CL Handshakes with D = DE, DF, DG, DH, DI, DJ, DK, DL Handshakes with E = EF, EG, EH, EI, EJ, EK, EL Handshakes with F = FG, FH, FI, FJ, FK, FL Handshakes with G = GH, GI, GJ, GK, GL Handshakes with H = HI, HJ, HK, HL Handshakes with I = IJ, IK, IL Handshakes with J = JK, JL Handshakes with K = KL ∴ Total no. of handshakes = 66 6. (A) : We have,
18 16 . + Riya + Students Students
\ Total number of students = 16 + 18 + 1 = 35 7. (C) 8. (B) : 6 cells around 1 cell (6 + 2) cells around 2 cells (8 + 2) cells around 3 cells ⇒ 10 cells around a row of 3 cells
11. (C) : Five lakhs one hundred and nine = 500000 + 100 + 9
Class 4
= 500109
12. (A) : Height of tallest child = 105 cm Height of shortest child = 60 cm ∴ Difference = (105 – 60) cm = 45 cm 13. (A) : Side of Hexagon = 20 cm ∴ Perimeter of Hexagon = (6 × 20) cm = 120 cm Also, perimeter of Hexagon = Perimeter of star = 120 cm ∴ 10 × side of star = 120 ⇒ side of star = 12 cm 14. (C) : We have, ×
= 100
= 10
So,
+
+
⇒ 2
= 120
+ 10 = 120
⇒
= 55 = 55
10 = 45
15. (A) : Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30 16. (A) : 8000 = 50 × 80 × Y 8000 = 4000 × Y or 8000 = 4000 × 2 ∴ Y=2 17. (C) : The required order is, 860 gm, 700 gm, 350 gm, 90 gm i.e., R, P, S, Q 18. (C) :
+
or 3 ×
9. (A) 10. (C) : 8 × 4 = 32 8 × 5 = 40
8 × 8 = 64 8 × ? = 56 8 × 7 = 56 ? = 7
or
+
= `960
= `960
= `320 ∴
+
+
= `720 17
⇒
2×
⇒
28. (C)
+ 320 = `720 2
⇒
29. (C) : Perimeter of field = 4 × 480 m = 1920 m \ Distance covered by Vishal = (5 × 1920) m = 9600 m
= `400 = `200
19. (D) : No. of parts = 12 No. of unshaded parts = 4 4 1 = ∴ Fraction = 12 3
30. (B) :
20. (D) : (8 : 15 + 1 hr 45 min.) = 10 : 00 ∴ Minute hand point on 12
31. (B) : Total no. of burgers = 726 No. of burgers in 1 packet = 3 ∴ No. of packets = 726 ÷ 3 = 242 Now, Cost of 1 packet = `2 ∴ Cost of 242 packets = ` (2 × 242) = `484
21. (D) : 89 = LXXXIX. 22. (B) : Weight of 1 packet = 180 g and weight of 3 boxes + weight of 1 packet = 360 g ⇒ 3 × weight of 1 box + 180 g = 360 g ⇒ 3 × weight of 1 box = 180 g ⇒ weight of 1 box = 60 g. 23. (B) 24. (D) : The product X is not divisible by 54 as it is not divisible by 9. 25. (C) : We have,
x – 2 + 6 + 3 = 12 x – 2 + 9 = 12 x = 5
32. (D) : No. of group of adults = 7 No. of adult in 1 group = 2 ∴ Total no. of adults = 7 × 2 = 14 ∴ Total no. of children = 200 – 14 = 186 Now, No. of groups of children = 3 ∴ No. of children in 1 group =
26. (A) : No. of chocolates sold by Henry and Jenny = 90 or No. of chocolates sold by Henry = 90 – no. of chocolates sold by Jenny ...(i) Also, Chocolates sold by Henry = 2 × chocolates sold by Jenny. \ 90 – chocolates sold by Jenny = 2 × chocolates sold by Jenny or 3 × chocolates sold by Jenny = 90 ∴ No. of chocolates sold by Jenny = 30 27. (B) : Capacity of small container = 380 ml Capacity of big container = 1250 ml ∴ Capacity of tank = (8 × 380 + 1 × 1250) ml = (3040 + 1250) ml = 4290 ml
1×2=2 2×2=4 4×2=8 8 × 2 = 16
33. (A)
186 = 62 3
34. (D) : Total no. of peaches = 80 No. of rotten peaches = 80 × 2 = 16 10
∴ No. of peaches left = 64 35. (C) : 1 goat has 4 legs ∴ 60 goats have (4 × 60) legs = 240 legs 1 deer also have 4 legs ∴ 30 deer have (30 × 4) legs = 120 legs Also, 1 child has 2 legs ∴ 10 children have 20 legs ∴ Total no. of legs = 240 + 120 + 20 = 380
vvv
18
IMO Work Book Solutions
