__Quantitative Aptitude Application Problems for IBPS, SBI, RBI, NICL and other bank and insurance Exams__**1. On a car, the absolute discount offered is proportional to the number of days in the month it is sold. The marked price of car is Rs. 550000. In the month of December, it was sold at a price of Rs. 340285. What can be the maximum difference in selling price of car in year 2016 (in Rs.)?**

A) 67550

B) 13530

C) 21040

D) 16510

E) 18000

**2. A rectangle, PQRS, has a length of 24 cm and a breadth of 18 cm. One of its diagonals, PR, is trisected at points M and N. With MN as side length, a square is drawn. What will be the area of this square (in square cm)?**

A) 75

B) 84

C) 100

D) 125

E) 216

**3. Matt invested a sum of Rs. 42000 in simple interest for 3 years at 14% per annum rate of interest. After 3 years, he withdraws the complete amount and invests in compound interest at 10% per annum rate of interest for 2 years. How much amount will he receive at the end?**

A) Rs. 72242.5

B) Rs. 71342.84

C) Rs. 70000

D) Rs. 72164.4

E) Rs. 75166.3

**4. Blue and yellow colors were mixed in a plate A in ratio 7:4 and in other plate B it was mixed in ratio of 3:5. In what ratio should colors from these plates be mixed so as to form a mixture in ratio of 4:3.**

A) 91:70

B) 70:91

C) 40:121

D) 121:40

E) None of these

**5. Dane has some money in the form of 5 rupee and 10 rupee coins. The total number of coins he has is 54. However, while counting the money, he counted three of his 5 rupee coins as 10 rupee coins by mistake. He calculated that he has Rs. 355. How many coins of Rs. 5 do he has?**

A) 30

B) 36

C) 40

D) 45

E) 50

**6. Jack is warming up for his football match. He increases the intensity of warm up successively in three rounds. For first round of 10 minutes, he runs at a speed T km/hr. For next 20 minutes round, he doubles his speed. For final round of 40 minutes, he doubles his second round speed. Find the average speed of Jack in three rounds.**

A) 2T km/hr

B) 2.4T km/hr

C) 3T km/hr

D) 2.7 km/hr

E) 3.5 km/hr

**7. In a school, the student to teacher ratio is 40 : 1. However, as per rules, the ratio should be at most 25 : 1. If the minimum number of more teachers required to achieve desired ratio is 30, then how many students are there in the school?**

A) 1200

B) 1500

C) 1800

D) 2000

E) 2400

**8. There are 10 chairs around a round table. Seven persons, A, B, C, D, E, F and G, want to sit of these seats. C, D and E are friends and want to sit together. In how many ways can they be seated?**

A) 2520

B) 5040

C) 10080

D) 7560

E) 1260

**9. While calculating the average cost price of 8 items, a shopkeeper reversed the digits of one of the values, as a result of which the average increased by 24.75. If the cost prices of all the 8 items are 3-digit numbers then find the difference between the unit digit and hundred digit of that number.**

A) 1

B) 2

C) 3

D) 4

E) None of these

**10. After every 6 days of work, Allan takes a day off from work on Sunday. On this day, Paul works instead of Allan. The efficiency of Allan is one fifth of that of Paul. Due to this, a work that started on Monday gets finished in 41 days. In how many Allan alone would have finished this work?**

A) 50

B) 51

C) 56

D) 61

E) 66

**Solutions:**

**1. B)**We know that 2016 is a leap year. February will have a minimum of 29 days, and maximum days in a month can be 31. December also has 31 days.

⇒ Maximum difference in selling price of car in 2016 will be
difference in selling prices in February and December(or any other month with
31 days).

Discount offered in
December = Marked price – Selling Price = Rs. 550000 – Rs. 340285 = Rs. 209715

December has 2 days more
than February, and the difference in selling prices(or equivalently difference
in discounts) will be proportional to this difference in number of days.

⇒ Maximum difference in selling prices / Discount in December
= 2/31

⇒ Maximum difference in selling prices = Rs. 209715 × (2/31) = Rs. 13530

**2. C)**Length of diagonal = √ (square of length + square of breadth) = √ (24× 24 + 18 × 18) = 30 cm

If diagonal PR is
trisected at M and N, then length of MN will be one third of PR.

⇒ MN = 30/3 cm = 10 cm

With MN as side length, a
square is drawn.

∴ Area of square = 10 cm × 10 cm = 100 square cm

**3. D)**Matt invested a sum of Rs. 42000 in simple interest for 3 years at 14% per annum rate of interest.

We know, Simple interest
= (Principal × Rate × Time)/100

⇒ Amount at the end of 3 years = Principal + Simple Interest =
42000 + (42000 × 14 × 3)/100 = Rs. 59640

After 3 years, he
withdraws the complete amount and invests in compound interest at 10% per annum
rate of interest for 2 years.

We know, Amount = P [1 +
(R/100)]

^{T}= 59640 x [1 + (10/100)]^{2}= 59640 x 1.21 = 72164.4
∴ Amount Matt will receive at end is Rs. 72164.4.

**4. D)**Let colors from these plates be mixed in ratio of x:y

∴ Blue color in resultant mixture due to plate A = 7x/11

Blue color in resultant
mixture due to plate b = 3y/8

Yellow color in resultant
mixture due to plate A = 4x/11

Yellow color in resultant
mixture due to plate B = 5y/8

Total quantity of mixture
= (x + y)

As per requirement we
require Blue and Yellow color in ratio of 4:3

∴Total Blue color = 4(x + y)/7

∴ Blue color due to plate A + Blue color due to plate B =
Total blue color

⇒ 7x/11 + 3y/8 = 4(x + y)/7

⇒ (56x + 33y) / 88 = (4x + 4y)/7

⇒ 392x + 231y = 352x + 352y

⇒ 392x – 352x = 352y – 231y

⇒ 40x = 121y

⇒ x/y = 121/40

⇒ x:y = 121:40

**5. C)**Suppose Dane has M coins of 5 rupees and N coins of 10 rupees.

Total number of coins =
54

⇒ M + N = 54 ------- (i)

While counting, Dane
counted three of his 5 rupee coins as 10 rupee coins.

⇒ He counted (M-3) coins of 5 rupees and (N+3) coins of 10
rupees.

Dane calculated that he
has Rs. 355

⇒ 5 × (M – 3) + 10 × (N + 3) = 355

⇒ 5M – 15 + 10N + 30
= 355

⇒ 5M + 10N = 340

⇒ M + 2N = 68 ------- (ii)

Multiply (i) by 2, and
subtract (ii) from it.

2(M + N) – (M + 2N) = 2 ×
54 – 68

M = 40

∴ Number of 5 rupees coins are 40.

**6. C)**As per given information, there are three phases in which Jack runs.

For phase I), Speed = T
km/hr,

Time taken = 10 minutes =
(10/60) hours = 1/6 hours

Distance covered = T
km/hr × 1/6 hr = (T/6) km

For phase II), in this
phase speed is doubled, Speed = 2T km/hr,

Time taken = 20 minutes =
(20/60) hours = 1/3 hours

Distance covered = 2T
km/hr × 1/3 hr = (2T/3) km

For phase III), in this
phase speed is double of speed in second round, Speed = 2 × 2T km/hr = 4T km/hr

Time taken = 40 minutes =
(40/60) hours = 2/3 hours

Distance covered = 4T
km/hr × 2/3 hr = (8T/3) km

From this,

Total distance covered =
(T/6) km + (2T/3) km + (8T/3) km = (7T/2) km

Total time taken = (1/6)
hr + (1/3) hr + (2/3) hr = (7/6) hr

Now, we know that,
Average Speed = (Total Distance Covered/Total Time Taken)

Average speed =
(7T/2)km/(7/6)hr=3T km/hr

∴ Average speed for entire warm up = 3T km/hr

**7. D)**Suppose there are 40T students and T teachers.

As per given condition,
if 30 more teachers are added, ratio of students to teachers becomes 25 : 1.

⇒ 40T/(T+ 30) = 25/1

⇒ 40T = 25T + 750

⇒ T = 750/15 = 50

∴ Number of students in school = 40T = 40 × 50 = 2000

**8. B)**7 persons want to sit around a round table having 10 chairs.

C, D and E want to sit
together.

⇒ They can be treated as one person but internally they can be
arrange in

^{3}P_{3}ways, i.e. 6 ways
⇒ Effectively, there are 5 persons A, B, CDE, F and G

C, D, E can sit on any 3
consecutive chairs.

Now, next person
can sit on any of remaining 7 chairs, and so on.

⇒ Total number of ways in which this can be done =

^{3}P_{3}× (1 × 7 × 6 × 5 × 4)
= 6 × 840 = 5040

**9. B)**Let the hundred’s place digit, ten’s place digit and unit digit of the correct number are ‘a’, ‘b’ and ‘c’ respectively, so the number mistakenly written as ‘cba’ in the place of ‘abc’

We know that, formula for
average-

A

_{E}= S_{E}/n_{E}where S_{E}= sum of entities; n_{E}= number of entities; A_{E}= Average of entities
Now, according to the
question,

Average increased by
24.75, hence sum increased by 24.75 × 8 = 198.

The reserved no. is cba.
Now,

⇒ cba – abc = 198

cba can be written as 100
c + 10b + a and

abc can be written as
100a + 10b + c, so

The difference = 100 c +
10b + a – 100 a – 10b - c = 198

⇒ 99 c – 99 a = 198

⇒ (c – a) 99 = 198

Therefore, c – a = 198/99
= 2

Hence the required
difference is 2

**10. D)**Suppose Allan takes T days to finish the work. So, in a day, Allan completed 1/T of total work.

The efficiency of Allan
is one fifth of that of Paul.

⇒ Paul completes 5/T of work in a day.

Starting with Monday,
work gets finished in 41 days.

41 days = 5 Sundays + 6
cycles from Monday to Saturday

⇒ Allan worked for 36 days, and Paul worked for 5 days.

⇒ 36/T + 5(5/T) = 1

⇒ T = 61

∴ Allan finishes work alone in 61 days.