Mock Test-3 (Paper-1) (Question 101-125)

Total Questions: 25

11. Let x be an odd natural number. If x is divided by 6, it leaves a remainder y. If y² is divided by 4, it leaves remainder of z. Which of the following must be true for z?

Correct Answer: (c) z = 1
Solution:x = 6Q + y

y² = 4Q¹ + z

The value of z may be 1, 2 or 3.

The value of y may be 1, 3, or 5 as if 2 or 4 be the value,

y² will be exactly divisible by 4.

= z = 1

12. Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy at least 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy?

Correct Answer: (c) Either 10 or 11
Solution:According to the question,

Ice-creams + Cookies + Pastries = 32

⇒ 9 + 11 + 12 = 32 or, 9 + 10 + 13 = 32 10 + 11 +12 ≠ 32

Hence, either she but 10 or 11 Cookies.

13. In how many ways a committee consisting of 3 men and 2 women, can be chosen from 7 men and 5 women?

Correct Answer: (b) 350
Solution:Out of 7 men, 3 men can be chosen in ⁷C₃ ways and out of 5 women, 2 women can be chosen in ⁵C₂ ways. Hence, the committee can be chosen in ⁷C₃ × ⁵C₂ = 350 ways.

14. Find the wrong term in the following number series.

114. 1, 14, 31, 64, 133, 264

Correct Answer: (c) 264
Solution:

15. Acylindrical can whose base is horizontal and is of internal radius 3.5 cm contains sufficient water so that when a solid sphere is placed inside, water just covers the sphere. The sphere fits in the can exactly. The depth of water in the can before the sphere was put, is

Correct Answer: (c) 7/3 cm
Solution:

Height of water after ball is immersed

= 3.5 × 2 = 7cm

⇒ Volume of water

= πr²h - 4/3 πr³

= πr²(h - 4/3 r)

= 22/7 × 3.5 × 3.5 (7 - 4/3 × 3.5)

= 11 × 3.5 (7/3) = 269.5/3 cm³

Volume of water before ball was immersed

= π (3.5)² × h = 269.5/3

= h = (269.5 × 7)/(3 × 3.5 × 3.5 × 22) = 7/3 cm

16. A and B can complete a piece of work in 80 days and 120 days respectively. They started working together but A left after 20 days. After another 12 days C joined B and they completed the remaining work in 28 more days. In how many days can C alone complete the work?

Correct Answer: (b) 112 days
Solution:Work done by A and B is 20 days

= 20(1/80 + 1/120) = 20((3 + 2)/240) = 5/12

Work done by B in 12 days = 12/120 = 1/10

Remaining work = 1 - 5/12 - 1/10 = (60 - 25 - 6)/60 = 29/60

Let C alone do the work in x days.

∴ 28/120 + 28/x = 29/60

⇒ 28/x = 29/60 - 28/120 = (58-28)/120 = 30/120

⇒ 28/x = 1/4 ⇒ x = 28 × 4 = 112 days

17. A, B and C walk 1 km. in 5 minutes, 8 minutes and 10 minutes respectively. C starts walking from a point at a certain time, B starts from the same point 1 minutes later and A starts from the same point 2 minutes later then C, then A meets B and C after.

Correct Answer: (a) 5/3 min, 2 min
Solution:Speed of A, B and C

= 1000/5, 1000/8, 1000/10

= 200 m/min.., 125m/min.., 100m/min.

Distance travelled by B and C before A starts

= 125, 200 metres

Time taken by A to meet B and C

= 125/(200-125), 200/(200-100) = 5/3 min, 2min

18. The interest received on a sum of money when invested in scheme A is equal to the interest received on the same sum of money when invested for 2 years in scheme B. Scheme A offers simple interest (p.c.p.a.) and scheme B offers compound interest (compounded annually). Both the schemes offer the same rate of interest. If the numerical value of the number of years for which the sum is invested in scheme A is same as the numerical value of the rate of interest offered by the same scheme, what is the rate of interest (p.c.p.a) offered by scheme A?

Correct Answer: (d) 2 2/99 %
Solution:Let the rate of S.I. be r % per annum.

∴ Time = r years

S.I. = (Principal × Time × Rate)/100

C.I. Principal[(1 + Rate/100)ᵀⁱᵐᵉ -1]

According to the question,

(P × r × r)/100 = P[(1 + r/100)² - 1]

(r ²)/100 = 1 + (r²)/10000 + (2r)/100 - 1

⇒ (r²)/100 = r/10000 + 1/50 ⇒ (r²)/100 - r/10000 = 1/50

⇒ (100r - r)/10000 = 1/50 ⇒ 99r = 10000/50 = 200

⇒ r = 200/99 = (2 2/99)% per annum

19. In a class of 168 students, boys and girls are in the ratio 5:7.50% of the total students can speak only Hindi. The ratio of number of students speaking only Hindi to that speaking only English is 21: 16. The ratio of boys speaking English only to that of girls speaking English only is 3: 5. If the number of boys speaking both English and Hindi is 12, what is the number of girls speaking Hindi only? (Assume that all students speak at least one language).

Correct Answer: (b) 50
Solution:Number of boys in class

(5/(5 + 7)) × 168 = 70

Number of girls = 168 - 70 = 98

Number of students speaking

Hindi only = 168/2 = 84

⇒ Number of students speaking English only

= 16/21 × 84 = 64

Number of girls speaking English only = 64 - 24 = 40

Number of students speaking both

= 168 - (84 + 64) = 20

Number of girls speaking both = 20 - 12 = 8

= Number of girls speaking only Hindi

= 98 - (40 + 8) = 50

20. The average of 5 consecutive numbers is n. If the next two numbers are also included, the average of the 7 numbers will

Correct Answer: (b) increase by 1
Solution:Let the numbers be n - 2, n - 1, n, n + 1 and n + 2.

Their average = n.

Next two consecutive numbers are n + 3 and n + 4

Therefore the average of 7 consecutive numbers

= ((n - 2) + (n - 1) + n + (n + 1) + (n + 2) + (n + 3) + (n + 4))/7

= (5n + 2n + 7)/7 = n + 1