Del enunciado
CRONOMETRÍA
Un campana campanario rio tarda tarda 6 segundos segundos en en tocar tocar 4 campanadas. Calcula el tiempo que tardará en tocar 8 campanadas
(24 − x) − x = 6
1.
A) 10 s D) 16 s
B) 12 s
18 = 2x
→
x = 9
Por lo tanto, son las 9 a.m.
C) 14 s E) 8 s Un reloj reloj se atrasa atrasa 4 min min cada cada hora. hora. Si hace hace 8 horas que se atrasa; determina la hora real cuando dicho reloj marca 10:00 a.m.
3.
–1 # Campanadas
# Interv Intervalo aloss
Tiempo
4 8
3 7
6 x
A) 10 h 20 min B) 10 h 24 min C) 10 h 32 min D) 10 h 16 min E) 11 h 10 min
O 3x = 7(6) → x = 14 C P E r o H Se atrasa f : P A C Por lo tanto, tardará 14 segundos. 4 min x
Determina qué hora es sí, en este instante, instante, el tiempo que falta para acabar el día excede en 6 horas al tiempo transcurrido.
2.
A) 1 p.m. D) 6 a.m.
B) 11 a.m.
Sabemos que
En 1h x = 4(8) = 32 min 8h HR
= HM + ATRASO
∴ Hora real = 10 h + 32min <> 10h 32 min
C) 9 a.m. E) 8 p.m.
Halla el ángulo formado por las manecillas del reloj cuando marca les 5:40.
4.
x
ora exacta
T. transcurrido
0h
24–x T. por por tran transc scur urrir rir
x
24h
A) 65° D) 70°
B) 60°
C) 80° E) 75° - 1 -
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¿A qué hora, exactamente, entre las tres y las cuatro, las manecillas de un reloj se deben superponer? 6.
12
9
3
5 : 40
Aplicando Aplicando
4 11
C) 3:15
D) 3:18
4 6
B) 3 : 16
A) 3:20
E) 3:22
5 12
11 θ = M − 30H 2 11 θ = (40) − 30(5) → θ = 70° 2
2 9
3:x
3
6
Determina a qué hora entre las 3 y las 4 las agujas de un reloj forman 24° por primera vez (cuando el minutero no pasa al horario).
5.
A) 3:16 D) 3:18
B) 3:26
C) 3:12 E) 3:34 P
9
3
Aplicando
Desarrollando
5
11 M + 30H 2 11 24 = − (x) + 30(3) 2 x = 12
θ=−
x = 16
→
4 11
E
3:x
4 6
O C
11 (x) = 30(3) 2
f : P A C H 4 Por lo tanto, se superponen a las 3 : 16 11
r o
12
Cuando las agujas están superpuesta s uperpuestass (θ = 0°), se 11 M = 30H , entonces cumple que 2
Deduce qué hora marca marca el reloj reloj mostrado mostrado en la figura. 7.
11 10
12
1 2
9
3 8
4
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Entonces 1 min
12
39 min 12
< > 6° 39 x = 12 (6) = 19,5° < > x
x/2 9
3
∴ α = 180° + 19,5° = 199,5°
6x
6
Del gráfico Para el horario
α+
Para el minutero
x = 30° 2 2α = 60° − x
…. (1)
2α + 6x = 180° …. (2)
ENTRETENIMIENTO 01
Reemplazando (1) en (2)
Un boxeador da 7 golpes en 5 segundos. ¿Cuánto tiempo demorará en dar 49 golpes?
9.
(60 − x) + 6x = 180 5x
= 120 →
x = 24
Por lo tanto, el reloj marca las 2:24.
8.
A) 35 s D) 40 s
O C
Un reloj señala las 6 h 39 min. Determina en C H E r o f : P A P
qué punto del cuadrante se encuentra el horario. A) 199,5° D) 205,5°
B) 189,5°
C) 209,5° E) 219,5°
B) 1 min 20 s
C) 25 s E) 60 s
–1
# Golpes
# Interv Intervalo aloss
Tiempo
7 49
6 48
5
6x = 48(5) → x = 40 Por lo tanto, demorará 40 segundos. 12
9
3
10
Arturo al observar un campanario campanario nota que
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Un anciano al caminar por la calle se da cuenta que da 5 golpes con su bastón en 8 segundos. ¿En cuántos segundos dará 10 golpes?
12.
–1 # Campanadas
# Interv Intervalo aloss
Tiempo
5 25
4 24
10
A) 16 s D) 24 s
B) 18 s
4x = 24(10) → x = 60
–1
Por lo tanto, demorará 60 segundos.
Un campanario tarda n 2x s en tocar tantas campanadas como n veces el tiempo que demora entre campanada y campanada. Halla el tiempo en función a n que demora entre campanada y campanada si es igual a x.
# Golpes
# Interv Intervalo aloss
Tiempo
5 10
4 9
8
11.
A) D)
n +1 s n 4n3
+1
2n
B)
n(n + 2) s n −1
C)
C) 20 s E) 12 s
4x = 9(8) 9(8) → x = 18 Por lo tanto, dará 10 golpes en 18 segundos.
Un campa campanar nario io toca toca tanta tantass campa campanad nadas as como horas indica. ¿Cuántas campanadas tocarán durante todo el primer medio día de hoy? n2 + 1 O s E) C P 2n E r o B) 78 C) 24 f : P A C H A) 68 D) 56 E) 38
s
ns
13.
Realizando un esquema n 2x
Hora 1
x
2
x
3
x
4
(nx (nx – 1) inter interva valos los
Donde
x(nx − 1) = n 2 x nx − 1 = n 2
nx
Campanadas
1h 2h 2 h 3h 3 h 4h 4 h …. 12 h ↓ ↓ ↓ ↓ ↓ 1 2 3 4 … 12
N° total de 12(13) ∴ = 78 = 1 + 2 + 3 + ... + 12 = 2 campanadas
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y B da 6 campanadas en 15 horas. ¿Cuántas horas horas transcur transcurren ren hast hastaa que que vuelven vuelven a tocar tocar simultáneamente?
Por lo tanto, demora 58 segundos.
A) 12 h D) 36 h
El campanario de una iglesia estuvo tocando durante 21 segundos, si se escucharon tantas campanadas como 10 veces el tiempo que hay entre campanada campanada y campanada. campanada. ¿Cuánto tiempo tiempo empleará este campanario para tocar 7 campanadas?
B) 21 h
C) 18 h E) 24 h
16.
Tiempo de Tiempo total = N° de intervalos cada int ervalo t A =
Campana A:
tB =
Campana B:
35 5 15 5
→
A) 7 s D) 6 s
B) 9 s
C) 8 s E) 10 s
t A = 7 h
Realizando un esquema
→
tB
Finalmente, ambos vuelven a simultáneamente cada: mcm(7;3) = 21 h
= 3h tocar
21 s 1
t
2
t
3
t
10t
4
(10t (10t – 1) inter interva valos los
Donde
t × (10 t − 1) = 21
Un reloj da 6 campanadas en 10 segundos. 10 t 2 − t − 21 = 0 O 2t − 3 = 0 Si un alumno resuelve un problema y escucha en C −3 2t P E → 3 r o demora H ese lapso 30 campanadas. ¿Qué tiempo 5t 7 t= f : P A C 2 en resolver el problema, problema, si la primera y la última campanada se escuchó en el instante que 3 Es decir, en 21 s se escucharon 10 = 15 empieza y termina de resolver el problema, 2 respectivamente? campanadas, luego 15.
A) 45 s D) 42 s
B) 60 s
–1
C) 58 s E) 50 s
–1 # Campanadas
# Interv Intervalo aloss
Tiempo
15 7
14 6
21 x
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Un reloj se adelanta 36 minutos cada 2 horas y otro se atrasa 30 minutos cada 5 horas. ¿Dentro de cuánto tiempo volverán a marcar la misma hora? 17.
A) 25 h D) 32 h
B) 28 h
C) 30 h E) 34 h
Se atrasa
En
3 min
40 min
×3
×3
9 min x Sabemos que
Reloj A : Se adelanta 36 min cada 2 h Sea Reloj B : Se adelanta 30 min cada 5 h
2h 8h
9(8) x = 2 = 36 min
HM = HR + ADELANTO
∴ Hora marcada = 2h 45 min+ 36 min = 3h 21 min
En una hora se separan hora exacta 6 min
18 min
atrasa
adelanta
Hace 12 horas y media se descompuso el reloj de la oficina, de modo que sufre un atraso de 8 min cada 4 h. Si en este instante dicho reloj marca las 9 h 57 min. ¿Cuál es la hora exacta?
19.
24 min min
Para que dos relojes marquen la misma hora, uno debe adelantar al otro 12h <> 720 min Adelanta
24 min 12h <> 720 min Donde
24x = 720 →
TiempoP r
o f :
1h x
A) 10 h 22 min B) 12 h 23 min C) 14 h 24 min D) 15 h 34 min E) 16 h 25 min O C E
H P A C
Se atrasa 8 min x
x = 30 h
En 4h 8(12,5) x = 4 = 25 min 12,5 h
∴ Hora real = 9h 57 min+ 25 min = 10h 22 min El reloj de la catedral se adelanta 3 minutos cada 40 minutos. Si este desperfecto ocurre ya hace 8 horas, ¿qué hora marcaran las agujas del reloj si la hora exacta, al cabo de las 8 horas, es
18.
Un reloj se retrasa 8 minutos cada 24 horas. Si este marca marca la hora correcta correcta 7 a.m. el 2 de 20.
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Faltan para las 8:00 a.m., la mitad de los minutos que pasaron desde las 6:00 a.m. de esta mañana, hasta la hora actual. ¿Qué hora indica el reloj? 22.
Del 2 de mayo (7 a.m.) al 7 de mayo (1 p.m.), transcurrieron 126 horas, entonces Se atrasa
En
8 min x
24 h 8(126) x = 24 = 42 min 126 h
Sabemos que
HM
A) 7:10 a.m. a.m. D) 7:40 a.m.
B) 7:20 a.m.
= HR − ATRASO ora exacta
2k
∴ Hora marcada = 1p.m. − 42 min = 12h 18 min
T. transcurrido
¿Qué hora es si hace 4 horas faltaba, para acabar el día, el triple del tiempo que faltará para acabar el día, pero dentro de 4 horas?
21.
k T. por por trans transcu curr rrir ir
x
6h
A) 16 h o 4 p.m. C) 18 h o 6 p.m. D) 19 h o 7 p.m.
C) 7:30 a.m. a.m. E) 7:45 a.m.
8h
2h <> 120 min
Del gráfico 3k
= 120
→
k = 40 h
∴ Hora exacta = 6 a.m. + 80min = 7:20 a.m.
B) 17 h o 5 p.m. E) 20 h o 8 p.m.
hora exacta
4h
4h x
2 O 3. Un reloj tiene 3 minutos de retraso y sigue C P E retrasándose a razón de 3 segundos por minuto. r o H f : P A C ¿Cuántos minutos deben transcurrir para tener una hora de retraso? k
24h
A) 1 230 min D) 1 840 min
B) 1 140 min min
C) 1 540 min E) 1 780 min
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Por lo tanto, deben trascurrir 1140 min. Se atrasa 3 min x
Tres relojes A, B y C se sincronizaron simultáneamente al mediodía. Si el reloj de A se atrasa 5 minutos por hora, el de B se adelante 5 minutos y el de C señala la hora correcta, ¿dentro de cuánto cuánto tiempo tiempo los los minuteros minuteros de los los 3 relojes relojes equidistarán entre sí? 24.
A) 4 h D) 8 h
B) 5 h
Sabemos que
En 1h x = 3(6) = 18 min 6h HR
= HM + ATRASO
∴ Hora real = 8 : 17 + 18 min = 8 : 35
C) 6 h E) 10 h Un reloj se atrasa 5 min cada 40 min, si ahora marca las 6 h 30 min y hace 8 horas que se atrasa, la hora correcta es: 26.
Sea “x” la cantidad de horas necesarias para que se cumpla la siguiente situación
A) 7:45 D) 7:15
5 min
C
B) 8:05
C) 7:30 E) 8:30
30° 120°
A
120° 120°
B P
r o
f :
H P A C
En cada hora el minutero B adelanta 5 min al minutero minutero C, es decir 30°, 30°, luego Tiempo
ngulo
Se atrasa O C 5 min E ×3
15 min x Sabemos que
En 40 min ×3
2h 8h HR
15(8) x = 2 = 60 min = HM + ATRASO
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Dos relojes son sincronizados a las 8:00 a.m. Uno de ellos se adelanta 15 segundos cada cuarto de hora y el otro se atrasa 45 segundos cada hora. ¿Cuántos minutos estarán separados a las 8:00 p.m. los minuteros de los 2 relojes?
29.
Desde las 6:00 a.m. hasta las 8:00 p.m. hay 14 horas, luego Se adelanta
En
75 s x
1h 14 h
Entonces
A) 22 min D) 24 min
B) 23 min
C) 21 min E) 25 min
x = 75(14) x = 1 050 s <> 17min 30s Analizando en una hora
Sabemos que
HM = HR + ADELANTO 45 s
∴ Hora marcada = 20h 17 min 30s
hora exacta
atraso
60 s adelanto
105 s
Un reloj se adelanta 1 minuto cada 15 minutos, si ahora marca las 04:20 y hace 8 horas que se adelanta, adelanta, la hora correcta es: 28.
A) 03:40 D) 03:10
B) 03:20
Se adelanta
En
1 min
15 min
×4
×4
4
1h
En una hora los relojes están separados 105 s, además desde las 8 am hasta las 8 pm hay 12 horas, es decir Separación
C) 03:48 E) 03:14
105 s O420 s <> 7 min C P E x r o H f : P A C Entonces
Tiempo
1h 4h 12 h
×4
4x = 7(12) → x = 21 min
Por lo tanto, estarán separados 21 minutos.
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19 días
11 días
En una día
13 julio 31 julio 1 agosto 11 agosto
hora exacta 3 min atraso
12:00
12: 00
12: 00
12:00
30 días
2 min adelanto
Por lo tanto, volverá a señalar la hora correcta el 11 de agosto, 12:00. 5 min
Para que ambos relojes coincidan, uno de ellos debe separarse del otro 12h <> 720 min Separación
5 min 720 min
Tiempo 1 día 720(1) x = 5 = 144 días x
Son más de las seis sin ser las ocho y hace diez minutos, los minutos que habían transcurrido desde las 6 eran iguales a 1/9 del tiempo que faltarían transcurrir hasta las ocho dentro de diez minutos. ¿Qué hora es? 32.
A) 5:20 D) 7:15
B) 6:25
C) 6:20 E) 8:35
Por lo tanto, cada 144 días ambos relojes coinciden. ora exacta
k 10 min min
Un reloj se atrasa un minuto por hora. Si O 6h C empieza empieza exactame exactamente nte el miércole miércoless 13 julio a P de julio E r o H las 12:00 h, ¿Cuándo volverá a señalar la f :hora P A C exacta? 31.
A) 11 de agosto: agosto: 12:00 h B) 12 de agosto agosto;; 12:00 12:00 h C) 09 de agosto agosto:: 09:00 09:00 h
10 min min
x
9k 8h
2h <> 120 120 min
Del esquema k + 10 + 10 + 9 k
= 120 10 k = 100 → k = 10 min
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¿Qué hora?, si dentro da 20 minutos el tiempo que faltará pera las 7:00 es el doble del tiempo transcurrido desde las 5:00 pero hasta hace 10 minutos. 35.
k
ora exacta
k
13 min min
7 min
x
11h
13h
A) 5:30 D) 6:50
B) 5:55
C) 5:45 E) 5:40
2h <> 120 120 min min
Del esquema k + 7 + 13 + k = 120 2k = 100 →
hora exacta
k = 50 min
∴ Hora exacta = 11h + 57 min = 11h 57min
k
20 min min
10 min min
2k
x
5:00
7:00
2h <> 120 120 min
Del esquema Ya pasaron las las 15:00 h sin ser las las 16:00 h. Si hubiera pasado 25 minutos más faltaría para las 17:00 h los mismos minutos que pasaron desde las 15:00 h hasta hace 15 minutos. ¿Qué hora es?
34.
A) 15:55 D) 15:20
B) 15:50
k
C) 15:15 E) 15:30
k + 10 + 20 + 2k
= 120 3k = 90 →
k = 30 min
∴ Hora exacta = 5:00 + 40 min = 5:40
O C un alumno de la academia le 3 6. A las 7:15 p.m. un P E r o f : P A C H dice a su compañera: “Cuando la suma de cifras de las horas transcurridas sea igual al doble de las ora horas que quedan por transcurrir, será la hora de exacta k 25 min min 15 min min salida”. ¿A qué hora terminan las clases en la academia?
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Del enunciado
a + b = 2(24 − ab)
12
a + b = 48 − 2(10a − b)
Ef ectuando y simplificando
7a + b = 16 ↓ ↓
9
1 9 2 2
Como el dialogo se realizó a las 7:15 p.m., se deduce a = 2 y b = 2
∴ Las clases terminaron a las 22 h <> 10 p.m.
6
3
min
Sabemos que 1 min <> 6°, entonces por regla de tres 1 min 7 min
¿Cuál es el menor ángulo que forman las manecillas de de un reloj cuando son las 12:58?
x
< > 6° x = 7(6) = 42° < > x
37.
¿Qué ángulo forman las manecillas de un reloj a las 04:20 p.m.? 39.
A) 234° D) 310°
B) 30°
C) 319° E) 41°
A) 10° o 350° B) 30° o 330° D) 120° o 240°
12
O
C) 150° o 210° 210° E) 20° o 340°
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Del gráfico 12
Para el horario horario Para el minutero
9
8:x
3 8
Reemplazando (1) en (2) 6x + x = 90 7x = 90 → x = 12
6
11 M 2 11 108 = 30(8) − (x) 2 x = 24 min
θ = 30H −
Aplicando
Desarrollando
Por lo tanto, tanto, formaran dicho dicho ángulo a las 8:24
41.
x → x = 2α …. (1) 2 6x + 2α = 90° …. (2)
α=
¿Qué hora indica el reloj de la figura? 11 10
12
1 2
Por lo tanto, son las 03 : 12
6 7
6 7
¿A qué hora entre las 3 y las 4 (el minutero ya paso al horario) el minutero y el horario formarán un ángulo que sea la cuarta parte del ángulo exterior? 42.
5 8 4 D) 04 : 30 3
A) 03 : 12
B) 03 : 29
4 3
5 11 4 E) 03 : 30 8
C) 03 : 29
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43.
¿Qué hora indica el reloj de la figura? 11
12
¿Qué hora marca el reloj de la figura mostrada, sabiendo además que θ − α = 7° ?
44.
1
10
11
2
9
3 4
8
5
6
2
9
4 7
7
5 7 2 D) 04 : 40 3
A) 03 : 34
B) 03 : 40
8 6
1
10
3 8
12
2 7 2 E) 02 : 28 6
C) 02 : 34
A) 04:24 D) 04:35
6
5
B) 04:25
C) 04:38 E) 04:40
Por dato θ − α = 7° 12
11 10
x/2 9
12
1 2
9
3 6x
3 8
O
x/2
6x
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¿Cuál es el menor ángulo (en grados) que forma el segundero con el horario cuando sean las 5h 16m 12s? 45.
A) 86,1° D) 93,5°
B) 91,5°
Como la hora observada adelanta 55 minutos a la verdadera, se deduce que el ángulo formado por el minutero y el horario es 30°
C) 92,6° E) 94,7°
12 30°
6x
30°
x/2 3
9
Si el minutero se desplaza “x” min, entonces ° x el horario recorre 2
6
Del gráfico
Sea θ el menor ángulo formado por el segundero y el horario 12
5h 16 min 12s S
9
θ H
18° 30° 30°
6x = 30° + 30° + 11x = 90 2
→
x + 30° 2 x = 16
4 min 11
Por lo tanto, la hora correcta es: 2 h 16
4 min 11
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Aplicando 12
x/2 2
11 M + 30H 2 11 40 = − (x) + 30(5) → 2
θ=−
x = 20
Por lo tanto, formaran dicho ángulo a las 5:20
3 6x
¿Qué hora indican las agujas del reloj?, si: 3α − θ = 40°
49.
Del gráfico
α = 30° +
Para el horario Para el minutero
x …. (1) 2
12
6x + α = 360° …. (2) 9
3
Reemplazando (1) en (2) 6x +
x + 30 = 360 2 13x = 330 2
6
→
x = 50
10 13
A) 10 : O
17 9 80
B) 10 :
97 8
C) 10 :
73 11 110
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Aplicando
Efectuando
11 M 2 11 260 = 30(10) − (x) 2 80 x = min 11
θ = 30H −
12
11x = 210 2
x + 180 2 x = 38
→
2 11
2 11
¿A qué hora entre 7 y la 8 las agujas de un reloj forman un ángulo de 80° por primera vez?
51.
¿Qué hora es?, si: α = θ 11
6x = 30 +
Por lo tanto, son las 10 : 38
80 Por lo tanto, son las 10 : 11
50.
Reemplazando (1) en (2)
1
7
7
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¿A qué hora entre la 1 y las 2, las manecillas de un reloj forman un ángulo de 137° por segunda vez? 52.
A) 1 h 46 min B) 2 h 46 min C) 3 h 46 min D) 4 h 45 min E) 2 h 35 min
12
3:x
9
x/2 6x 6
Del gráfico 6x = 90° + 12
11x = 180 2
137°
9
3
1:x
x + 90° 2
→
x = 32
8 min 11
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Del gráfico
• Regresa a su casa a las 2 : x 6x =
x + 30° 2
11x = 30 2
11
→
x = 5
5 min 11
12
10 9
Por lo tanto, forman ángulo llano cada 5 1h 5 min 11
6
Aplicando Cristina, Cristina, sale sale de su casa cuando cuando entre entre las 15:00 h y las 16:00 h, las agujas del reloj forman un ángulo de 180° y regresa cuando entre las 22:00 h y las 23:00 h las agujas del reloj se
55.
3
11 M = 30H 2 11 600 (x) = 30(10) → x = min 2 11 600
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Aplicando
11 M − 30H 2 11 x = (x) − 30(3) 2 9x 90 = → x = 20 min 2
θ=
IV. EI tercer tercer jueves jueves es igual igual a la suma suma del del primer primer viernes y el segundo jueves. V. La vísp víspera era del del aniv anivers ersari arioo cae vier viernes. nes. A) FFFVV D) VVFFV
B) FVVVF
C) VFVVV E) FFVVV
Por lo tanto tanto,, son las 3:20 1 Calendarios
En un mes mes hay 5 jueves, jueves, 5 viernes. viernes. Halla Halla que fecha cae el tercer sábado de dicho mes.
57.
… … … … … … …
… … 8 … … 15 … … 22 … … … 29
Aniversario
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Del enunciado
100 días
jueves Hoy
<>7+2
… –1 +1 +1 +1 es lu lunes <> +1 es lunes <> mañana es lunes → hoy es domingo
x
Del esquema se deduce que dentro de 100 días equivale a avanzar 2 días Por lo tanto, dentro de 2 días será sábado.
Piden
… –1 +1 < > hoy
Por lo tanto, hoy es domingo.
Entretenimiento
El 14 de febrero de 1996 fue miércoles, ¿Qué día cayó el 14 de febrero de 1999?
60.
62
El año 2000 es bisiesto. Si Miraval cumplió
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Un Operario trabaja 10 días seguidos y descansa dos. dos. Si empieza a trabajar un día lunes. ¿Cuántos días han de pasar para que le toque descansar sábado y domingo? 63.
A) 96 D) 94
B) 82
C) 85 E) 102