RBI Grade B 2018 Maths Questions Quiz, PDF With Explanation

Direction : Study the following questions carefully and choose the right answer.
Important for :
1
Sourav Ganguly wants to buy a total of 100 sports equipment using exactly a sum of Rs.1000. He can buy ball at Rs.20 per unit, wicket at Rs.5 per unit and bat at Rs.1 per unit. If he has to buy at least one of each equipment and cannot buy any other type of equipment, then in how many distinct ways can he make his purchase?
» Explain it
C
Let the number of ball, wickets and bats purchased be A, B and C, respectively.
 
Thus,
 
20A + 5B + C = 1000 and A + B + C = 100
 
Solving the above two equations by eliminating C, we Get 
 
19A + 4B = 900
 
⇒ B = 225 –  19 A
4
Now, as B is the number of wickets and 0 < B < 99,
 
So, putting these limiting values of B in the above equation will provide the value of A as 27 < A < 47.
 
Since A has to be the multiple of 4, so possible values of A are 28, 32, 36, 40 and 44.
 
Now, for A = 28 and 32; A + B > 100, so these values of A can be rejected.
 
For all other values of A, we get the desired solution:
 
A = 36, B = 54, C = 10
 
A = 40, B = 35, C = 25
 
A = 44, B = 16, C = 40
 
Thus, there are three possible solutions.

Hence, option (C) is correct.
 
2
The table below shows the distribution of the students in different schools in Pune.  

School Percentage
DPS 31.25
PPS 12.50
DAV 18.75
KV 15.00
RPS 10.00
Carmel 12.50

If the number of students in DAV, DPS and PPS is 250 and the ratio of the number of girls to boys in DPS is 3 : 2, find the number of girls in DPS.
» Explain it
D
Percentage of students, in DAV, DPS and PPS = 31.25 + 12.50 + 18.75 = 62.5%

This is equal to 250

⇒ Total strength of the school =  100  × 250
62.5

8  (250) = 400
5

∴ The number of students in DPS
 
31.25  × 400 = 125 students
100

Ratio of number of girls to boys = 3 : 2
∴ The number of girls in DPS =  3  × 125 = 75
5

Hence, option (D) is correct.

3
In the following figure, Δ ABC is a right angled triangle, right angled at B, intersecting the circle at E, B and D. AD is the angle bisector of ∠ BAC.
 
AB = 72 cm and AC = 75 cm. Also, BE and BD are two equal chords of the circle. What is the area of the given circle?
» Explain it
C
As AD is the angle bisector, it divides BC in the same ratio as AB to AC.

∴  BD  =  72  =  24
DC 75 25

∴  BD  =  24
BC 49

As Δ ABC is a right angled triangle, using the Pythagoras Theorem, ∴ BC = 21 cm
∴  BD =  24  × 21 =  72  cm
49 7

Also, BD = BE =  72  cm
7

Δ DBE is an isosceles right angled triangle.
∴ DE = 
72 2
7
As ∠ABC is right angle, clearly DE is the diameter of the circle.
=> Area of the circle = πr2
∴ The area of given circle = π  (
72 2
) 2
7 × 2  

2592 π
49

Hence, option (C) is correct.

4
Medha goes to a stationary shop and buy 4 pink sketch pens and some yellow sketch pens. The price of a pink sketch pen was twice that of the yellow sketch pen. When shopkeeper calculated the amount, it was found that the number of sketch pens of two colors had been interchanged. This increase the amount by 50%. What is the difference between the number of pink sketch pens to the number of yellow sketch pens in the original order? 
» Explain it
A
Let the number of yellow sketch pens be 'x'

The number of pink sketch pens is 4

Let the cost per yellow sketch pens be P

Then, the cost per pink sketch pens is 2P

Original cost = P × x + 4 × 2P = P(x + 8)

Cost after the colors had been interchanged = x × 2P + 4 × P = P (2x + 4)

It is given that this cost is 50% more than the original cost

Hence, P(2x + 4) = 1.5 [P(x + 8)]

⇒ 2x + 4 = 1.5x + 12

⇒ 0.5x = 8

⇒ x = 16

Therefore, the required difference = 16 – 4 = 12 Hence, option (A) is correct.
 
5
Ganguly plays a game wherein he tosses a coin a certain number of times. Each time the coin turns up heads, Ganguly is paid Rs. 385 and each time it turns up tails, he has to pay Rs. 150. Ganguly wins a net amount of Rs. 2030 by tossing the coin ‘n’ times. If the total number of tosses is less than 30, find the number of times the coin was tossed.
» Explain it
C
Let x be the number of times the coin turned up heads and y be the number of times the coin turns up tails.
 
According to question,
 
385x – 150y = 2030 ⇒ 77x – 30y = 406 ….. (i)
 
Since 77x and 406 are multiples of 7 so, 30 y should be a multiple of 7 or y itself is a multiple of 7
 
Let y = 7z
 
In (i)
 
77x – 30 × 7z = 406 ⇒ 11x – 30z = 58 …….. (ii)
 
By hits and trial,

If z = 1 then y = 7 × 1 = 7 and x =  58 + 30  = 8
11

If z = 12 then y = 7 × 12 = 84 and x =  58 + 360  = 38
11

If z = 23 then y = 7 × 23 = 161 and x =  58 + 690  = 68 and so on
11
 
But given that the total number of tosses is less than 30
 
i.e, n = x + y < 30
 
then only one case is valid, which is x = 8 and y = 7
 
Thus, the total number of tosses = 8 + 7 = 15

Hence, option (C) is correct.