NUMBERS & FUNDAMENTALS OF MATHS (PART-II)
NUMBERS & FUNDAMENTALS OF MATHS (PART-II)
QUERY 11
In Rs 15 how many chocolates you can buy if rate of 1 chocolate is Re 1 and you can also buy 1 chocolate in exchange of 3 wrappers?
A) 15
B) 18
C) 17
D) 22
MAHA GUPTA
15 Rs———-15 chocolates—-15 wrappers (5 set of 3 wrappers)
15 wrappers—5 chocolates—–5 wrappers (1 set of 3 wrappers + 2)
5 wrappers—–1 chocolate——1 wrapper + 2 already in stock
3 wrappers—–1 chocolate——1 wrapper
In the end you are left with 1 wrapper for which you can’t buy any more as you need atleast 3 ones.
So the number of chocolates you can buy in actual = 22 (option ‘D’)
QUERY 12
How many divisors are of 96?
A) 12
C) 15
B) 10
C) 14
MAHA GUPTA
1 x 96 = 96
2 x 48 = 96
3 x 32 = 96
4 x 24 = 96
6 x 16 = 96
8 x 12 = 96
You see 9 is not, 10 is not, 11 is not, 12 has come already.
When a factor has already appeared in the calculation stop yourself. Count all the factors thus found, and that is the answer. So the answer here is 12 (option ‘A’)
QUERY 13
A bag contains coins of Re 1, Rs 2 and Rs 5 denomination in the ratio of 2 : 3 : 4. If the total amount is Rs 280, find the number of coins of Rs 5 denomination in the bag.
A) 20
B) 30
C) 35
D) 40
MAHA GUPTA
In such sums always find the following two things:
i) Ratio of value of each coin, which can be found multiplying each denomination by its corresponding ratio
ii) Value of each denomination of coins
So ratio of value of each coin = (1*2) : (2*3) : (5*4) = 2 : 6: 20 = 1: 3 : 10
Now the sum of ratio = 14
But the sum given is 280
So the value of Rs 5 coins = (10/14)*280 = 200
Hence the number of Rs 5 coins = 200/5 = 40 (option ‘D’)
QUERY 14
A bag contains 50 p, Re 1 and Rs 2 coins in the ratio 2 : 3 : 4. If the total amount is Rs 240 what is the total number of coins?
A) 200
B) 150
C) 180
D) 90
MAHA GUPTA
In such a sum always find the value of each denomination and total number of coins according to the ratio given and add them.
Value of 50 p denomination here = (1/2)*2 = Re 1
Value of 1 Re denomination = 1*3 = Rs 3
Value of Rs 2 denomination = 2*4 = Rs 8
So the total value according to the ratio = Rs 12
And the number of coins according to the ratio = 2+3+4 = 9
But the total value in actual = Rs 240
If the value is 12 the number of coins = 9
If the value is 240 the number of coins = (9/12)*240 = 180 (option ‘C’)
QUERY 15
The ratio between the difference of denominator and numerator and the sum of denominator and numerator of a fraction is 2 : 9. What is the fraction?
A) 2/9
B) 5/6
C) 8/9
D) 7/11
MAHA GUPTA
Let the fraction = x/y
Now according to the question
(y-x)/(y+x) = 2/9
=> 2y + 2x = 9y – 9x
=> 7y – 11x = 0
=> x/y = 7/11
So the answer should be 7/11 (option ‘D’)
TRICK
Better to do this sum by answer options.
QUERY 16
A man fill a basket with eggs in such a way that every minute he adds same number of eggs already present in the basket. This way the basket gets completely filled in 60 min. After how many minutes the basket was 1/4th full?
A) 15 minutes
B) 20 minutes
C) 30 minutes
D) 58 minutes
MAHA GUPTA
Full filled—————–60 minutes
1/2 filled—————–60 – 1 = 59 minutes
1/4 filled—————–59 – 1 = 58 minutes (option ‘D’)
QUERY 17
A yearly payment to a servant is Rs 90 plus one turban. The servant leaves the job after 9 months and receives Rs 65 and a turban. Then find the price of the turban.
A) Rs 8
B) Rs 10
C) Rs 15
D) Rs 20
MAHA GUPTA
Let the price of turban Rs x
Now according to the question (x+90)/12 = (x+65)/9
=> x = 10
Therefore the price of the turban = Rs 10 (option ‘B’)
ANOTHER METHOD
Algebraic way at times proves to be more time consuming. We can also think the solution to this question like this:
Well, if the payment to be made in 9 months is considered exactly in terms of rupees and turban separately, the servant should get (90/12)*9 i.e. Rs 67.50 + (turban/12)*9 i.e. 3/4 of the turban. But he is getting rupees less by 67.50 – 65 = 2.50 and the whole of the turban.
Means this Rs 2.50 is 1/4 of the price of the turban.
Therefore the price of the turban = 4*2.50 = Rs 10 (option ‘B’)
QUERY 18
In a post-office, stamps of only Rs 7, Rs 8 and Rs 10 are available. Which amount of the below you can’t purchase these ones with?
A) 19
B) 20
C) 23
D) 29
MAHA GUPTA
Just see which multiple of the given denominations giving you the numbers given in answer options after adding any of the the given denominations
Let’s see
7*3 + 8 = 29 (so we can purchase 3 stamps worth Rs 7 and one stamp worth Rs 8 with 29)
8*2 + 7 = 23 (so we can purchase 2 stamps worth Rs 8 and one stamp worth Rs 7 with 23)
10*2 = 20 (so we can purchase 2 stamps worth Rs 10 with 20)
7*2 + any denomination is not making any number out of the given options
We are left with 19 only, so it’s our answer (option ‘A’)
QUERY 19
The sum of the digits of a two digit number is 10. If 18 be subtracted from it, the digits in the resulting number will be equal. What is the number?
A) 82
B) 91
C) 73
D) 64
MAHA GUPTA
Let the digit at unit place be x
and the digit at ten’s place be y
Therefore the number is 10y + x
And sum of the digits = x + y = 10
The number after 18 being subtracted from it = 10y + x – 18
but the digits get equal after 18 being subtracted from the original number
Means x = y
=> x = y = 5
Therefore 10y + x – 18 = 10x + x
=> 10y + x – 18 = 55
=> 10y + x = 73
But 10y + x is the original number —–shown above
Hence the answer is 73 (option ‘C’)
TRICK
Subtract 18 from all the given answers one by one, and see when you get same digits at unit’s and ten’s places. That option will be your answer.
Here 73 – 18 = 55, so 73 (option ‘C) is the answer.
QUERY 20
The sum of a two digit number and the number obtained by reversing its digits is a square number. How many such numbers are there?
A) 9
B) 8
C) 7
D) 6
MAHA GUPTA
Let the unit digit of the initial number be ‘x’ and the ten’s digit be ‘y’.
Therefore the number = 10y + x
And the number after the digits being reversed = 10x + y
Hence their sum = 11x + 11y
= 11(x +y)
But according to the question it’s a square number
If you see 11 is one of the two factors of the new number
Hence the number to be a square the other factor must be 11 also.
So x + y = 11
Obviously ‘1’ can’t be any of the two digit; and other pairs satisfying this conditions are 29, 38, 47, 56, 65, 74, 83, and 92
So we have 8 such numbers that are required. (option ‘B’)