Olympiad Test: Arithmetical Reasoning - Class 10 MCQ

Let Ram’s age = x, Mohan’s age = y, Lokesh’s age = z
∴ x = 3y …(1)
and z – 4 = 2 (x – 4)
z = 2x – 8 + 4 = 2x – 4 …(2)
x + 4 = 31 ⇒ x = 27
3y = 27 ⇒ y = 9
Using eq. (2) z = 2x – 4 = 2 × 27 – 4 = 50
Hence z – y = 50 – 9 = 41

Let Rajesh solves x sums correctly
Right sums = x, wrong sums = 2x
∴ x + 2x = 84
⇒ 3x = 84 
⇒ x = 28 

Let Punam’s age be 10x + y
∴ Her husband’s age = 10y + x
Now
 10y + x – 10x – y = 1/11 (10y + x + 10x + y)
⇒ 
⇒ 
⇒ 9y – 9x = x + y
⇒ 9y – y = x + 9x
⇒ 8y = 10x
⇒ 
If y = 5, then x = 4
Hence, Punam’s age = 10 × 4 + 5 = 45 years.

Let the number of 25 paise coins be x
No. of 20 paise coins = 324 – x
∴ 0.20 (324 – x) + 0.25 × x = 71 
⇒ 64.8 – 0.20x + 0.25x = 71 
⇒ 0.05x = 71 – 64.8 
⇒ 

Let the daughter’s age be x
Father’s age = 3x
Mother’s age = 3x – 9
Son’s age = x + 7
∴  
⇒ 2x + 14 = 3x – 9 ⇒ x = 23
Mother’s age = 3x – 9 = 3 × 23 – 9 = 60 years

Let the number of cows be x & no. of hens be y
As per the question,
4x + 2y = 2 (x + y) + 14
⇒ 4x + 2y = 2x + 2y + 14 
⇒ 2x = 14 ⇒ x = 7 

If I keep 2 or 3 or 4 in a pack, 
LCM of 2, 3, and 4 = 12
Then, the minimum number of sweets could be 12 + 1 = 13 so that 1 sweet left after packing.
But, if I keep 5 in a pack, I am left with none, (13 is not divisible by 5) then we have to take, 
12 × 2 = 24 
We have to add one so that 1 sweet left after packing in 2 or 3 or 4 groups and when in group of 5 then no sweets left. Therefore,
Total minimum sweets = 24 + 1 = 25 (25 is divisible by 5).
Hence, 25 is the correct answer.

Let the no. of adults = 2x
No. of children = 3x
Literate population
= (100 – 40)% of 2x + 85% of 3x
= 60% of 2x + 85% of 3x


Literate percent = 

 Given B – 40 = ⇒ 2B – 80 = C …(1) 
and C – 40 = B …(2) 
and C – 40 = A + 40 …(3) 
From (1) & (2) 

C = B + 40 = 120 + 40 = 160 
160 – 40 = A + 40 ⇒ A = 160 – 80 = 80 
7 + B + C = 80 + 120 + 160 = 360

To find the time in which all the bells ring together, we need to find the LCM of 36, 40, 48
Prime factorization of 36 = 2 × 2 × 3 × 3
Prime factorization of 40 = 2 × 2 × 2 × 5
Prime factorization of 48 = 2 × 2 × 2 × 2 × 3
Hence, LCM of  36, 40, 48 = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720 seconds.
∴ 

Let the number be 100
100 + 10% of 100 = 100 + 10 = 110
and 110 – 10% of 110 = 100 – 
= 110 – 11 = 99 

Cost price of 11 objects = Rs. 10
Cost price of 1 object = Rs. 
Selling price of 10 objects = Rs. 11
Selling price of 1 object = Rs. 
∴ Gain percent = 

Average monthly income

Income of that person = 120% of 5050 

= 12 × 505 = Rs. 6060 

Total no. of students = 200
Let the no. of sweets = x
∴ 
⇒ 
⇒ 200x – 150x = 4 × 150 × 200 
⇒ 

Let Rakesh paid Rs. x. 
Jay paid = x/2
Arjun paid = 
∴ Total bill = 

Part of Jay =