ARITHMETICMATHSNumber, HCF/LCM, Finding of Unit Digit, Sum of Series, etc.




Out of 100 families in the neighbourhood 50 have radios, 75 have TVs and 25 have VCRs. Only 10 families have all the three and each VCR owner has also TV. If 25 families have radio only, how many have only TVs?

A) 33
B) 25
C) 35
D) 55

Vaibhav Vats
Answer is 35 (option ‘C’)


Which least number shall be added to 8115 to make it a perfect square ?

A) 349
B) 166
C) 144
D) 194

Naveen M
I HOPE U KNOW SHORT CUT for finding √8115……it gives 90 as
quotient and 15 as remainder. the nearest to it is 91. Now find its square and subtract the
given number from it.

Thus 91 x 91 – 8115= 8281 ­- 8115 = 166. So option ‘B’ is the answer


The least positive integer that should be subtracted from 3011 X 3012 so that difference is a perfect square?

A) 3000
B) 3012
C) 3011
D) 3175

3011 × 3012 = 3011 × (3011 + 1)
= 3011² + 3011

If we subtract 3011 (here it’s the least positive integer to be subtracted) from the above we see that the remaining part is a perfect square (option ‘C’)


If the digits in the unit and the ten’s places of three digit number are interchanged a new number is formed which is greater than the original number by 63. Suppose the digit in the unit place of the original number be ‘x’; then all the possible value of ‘x’ are ?

A) (7, 8, 9)
B) (2, 7, 9)
C) (0, 1, 2)
D) (1, 2, 8)

Let the number is 100a + 10b + x
Therefore the number after the digits being interchanged according to the question 100a + 10x + b

Thus 100a + 10x + b = (100a + 10b + x) + 63
=> 10x + b = 10b + x + 63                     (By cancelling of 100a from both sides)
=> x = 7 + b

This implies that x > 7
But x must be smaller than 9 as no digit can be greater than 9

So x = (7, 8, 9) [option ‘A’]


If a natural number multiplied by 18 and another by 21 and the products thus got are added. Which one of the following could be the sum?

A) 2007
B) 2008
C) 2006
D) 2002

vaibhav jindal
Remember if a number is sum of product of two or more numbers it’s always divisible from the common prime factors of those numbers.

Here we see that 3 is a common prime factor of both 18 and 21, so the sum of the product must b divisible by 3.
We see that in the answer options the number 2007 only is divisible by 3.

So answer is 2007 (option ‘A’)


In a bag for every 2 notes of Rs 100 there are 5 notes of Rs 50, for every 3 notes of Rs 50 there are 4 notes of Rs 10. If the value of Rs 50 notes is Rs 2250, find the total value of money in the bag?

A) 4560
B) 4650
C) 5000
D) 5650

Number of Rs 50 notes = 2250/50 = 45
According to the question the number of Rs 100 notes = 45 × 2/5 = 18
And the number of Rs 10 notes = 45 × 4/3 = 60

Therefore the total value of money in the bag = (100 × 18) + 2250 + (60 × 10)
= Rs 4650 (option ‘B’)


The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels. What is the largest number of bananas that can be delivered to the market?

A) 525
B) 467
C) 500
D) 533

To find the largest number of bananas to be carried by it we’ll follow the following approach, nahin to saare kele wo oont hi kha jaayega.

The camel carries 1000 bananas first. Now better to find the number of kilometers it can walk eating all of those 1000 bananas and carry rest of the bananas i.e. 2000 in number to the point of that last kilometer. It eats 1 banana for every kilometer, means for those 1000 bananas to finish it’ll walk 1000 kilometer. To take the left over bananas it need to make 5 trips (3 TO and 2 FRO). So it’ll cover 200 km (1000/5) to take those 2000 bananas to the 200 km mark.

Now we have 2000 bananas in all to carry to the market at 200 km mark. Well, again we ‘ll have to apply the same approach. for those 2000 bananas it’ll make 3 trips (2 TO and 1 FRO). So to finish the next lot of 1000 bananas it’ll walk 333 km (1000/3) more. In this way we’ll have 1000 bananas at 533 (200+333) km mark.

Now we are left with 1000 bananas only to carry to the market and 467 (1000 – 533) km. Obviously it’ll eat 467 more of the bananas.

Hence the the largest number of bananas to be delivered to the market = 1000 – 467 = 533 (option ‘D’)


The half life period of an isotope is 2 hours. After 6 hours what fraction of the initial quantity of the isotope will be left behind?

A) 1/6
B) 1/3
C) 1/8
D) 1/4

Maverick Baloo Praful Nigam
Remains of the isotope left in first two hours = 1/2
So in next two hours the remains left = 1/2 of 1/2 = 1/4
And in next two hours the remains left = 1/2 of 1/4 = 1/8

Hence after 6 hours fraction of the initial quantity of the isotope left = 1/8 (option ‘C’)


A woman sells to the first customer half her stock of apples and half an apple, to the second customer half an apple and half of her remaining stock and so also to a third customer and to a fourth customer. She finds that she has now 15 apples left. How many apples had she at first?

A) 150
B) 180
C) 200
D) 250

Here the language of the question has to be noted carefully. According to this she sells to the third customer half her stock of her apples that left with her after making sales to the second customer and then half an apple from the remaining ones; but to the fourth half an apple first and then half of her remaining stock. Now see the solution:

Start such questions from the rear end.
After the woman gave half an apple and half of remaining stock to the 4th customer she had 15 apples left with her.
So before the fourth customer she had apples = (0.5 15*2) = 30.5

And before the 3rd customer she had = (30.5 + 0.5)*2 = 62

And before the 2nd customer she had (0.5 + 62*2) = 124.5

And before the 1st customer she had = (124.5 +0.5)*2= 250

So initially she had a stock of 250 apples. (option ‘D’)


A three digit number which when subtracted from another three digits number consisting of the same three digits in reverse order gives 594. the minimum possible sum of all the three digits of this number?

A) 15
B) 9
C) 8
B) 6

Let the initial 3 digit number = 100z + 10y + x
Therefore the number when the digits are reversed = 100x + 10y +z

Now according to the question:
100x + 10y + z – 594 = 100z + 10y +x

=> 99x – 99z = 594

=> x = 6 + z

x can’t be zero as the the number will become a two digit number when the digits are reversed; also y can be any digit.

Now according to the equation above the combination of digits at the unit’s & hundred’s places: 7 & 1, 8 & 2, 9 & 3

So the initial number could be any of 107 to 197 or 208 to 298 or 309 to 399

But we need the minimum possible sum of the digits
So the initial number is 107

Hence the minimum possible sum of the digits = 8 (option ‘C’)

NOTE: Better to do this question orally mixing hit and trial.

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Maha Gupta

Maha Gupta

Founder of and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)