Solution of Lekhpal 2015 Exam Paper Continues Part 3rd

Solution of Lekhpal Exam 2015 continues......

Q18. The height of an equilateral triangle whose each side is 4 cm, is

(a) 4√3 cm       (b) √3 cm      (c) 2√3 cm          (d) 3√3 cm.

Sol. As per the conceptual formula for the height of the equilateral triangle we know

Height (h) = (√3/2) × a       where “a” is the side of the equilateral triangle

Given data side of the equilateral triangle “a” = 4 cm.

Putting the given data in the formula we get the required answer

Height (h) = 2√3 cm.

Option (c)

Q19. A, B and C are shareholders in a company. In a certain year A got 1/3^rd share of the profit, B received 1/4^th share and C got Rs. 5000/-. Then how much profit did A get?

(a) Rs. 5000/-      (b) Rs. 1000/-   (c) Rs. 4000/-        (d) Rs. 3000/-.

Sol. Share of “A” in the profit in a certain year = 1/3^rd

Share of “B” in the profit in a certain year = 1/4^th

Share of “C” in the profit in a certain year = [1 – {(1/3) + (1/4)}] = 5/12^th

But according to the question “C” received as his share = Rs. 5000/-

Let the total profit in the company = x

Conceptually we know that (5/12) of x = 5000

→ x = 12000.

Now profit of “A” = (1/3) of 12000 = Rs. 4000/-

Option (c).

Q20. The smallest number by which 512 is multiplied so that the product is a perfect cube is

(a) 7                       (b) 8                    (c) 9                       (d) 10.

 Sol. The prime factors of 512 = 2×2×2×2×2×2×2×2×2

→ 512 itself a perfect cube of 8. So the smallest number by which 512 is multiplied so that the product is a perfect cube number is only and only 1, which is not preset in the given options. In such case it is suggested to each and every student that they leave this question number blank in their answer sheet which will represent your concept clearance.

Option (Do not mark any option leave it blank. This question is wrong.)

Q21. 720 sweets were distributed equally among the children in such a way that the number of sweets received by each child is 20% of the total number of children. How many sweets did each child receive?

(a) 14                    (b) 11                   (c) 15                   (d) 12.

Sol.  Let the total number of children = x

According to the question number of sweets received by each child = 20% of x = (x/5)

Total number of sweets to be distributed = 720  sweets.

Now (x) × (x/5) = 720

On solving we get x = 60 children.

So the number of sweets received by each child = (60/5) = 12 sweets.

Option (d)

Q22. Three different containers contain different quantities of a mixture of milk and water, whose measurements are 403 litres, 434 litres, and 465 litres. What biggest measure (in litres) must be there to measure all the different quantities exactly?

(a) 51                    (b) 41                    (c) 31                  (d) 70.

Sol. Given quantities 403 litres, 434 litres, 465 litres to be measured.

To solve such questions we have to just find the HCF of all given quantities.

Method (Trick) for calculating HCF :-

Prime Factors of smallest quantities from the given quantities:-

403 = 31 × 13 (among these two factors after checking we find 31 is the only common factor for other two remaining quantities.)

Hence the biggest measure (in litres) to measure all the different quantities exactly = 31 litres

Option (c)

Q23. If 1 is added to the denominator of a fraction, the fraction becomes ½. If 1 is added to the numerator, the fraction becomes 1. The sum of numerator and denominator of the fraction is

(a) 14                   (b) 21                     (c) 5                    (d) 11.

Sol. Conceptual Method :- Let the fraction be (x/y)

Where numerator is x and denominator is y.

According to the question:- 1 is added to denominator the fraction becomes ½ .

[x/(y + 1)] = ½

On simplifying we get the first equation 2x – y = 1   ........................ eq.no.1

According to the question:- 1 is added to numerator the fraction becomes 1 .

[(x + 1)/y] = 1

On simplifying we get the second equation x – y = -1   ........................ eq.no.2

On (eq.no.1 – eq.no.2) we get x = 2

From eq.no.1 (on putting x = 2) we get y = 3

Hence the faction is (2/3) and the required sum of numerator and denominator = 2 + 3 = 5.

Option (c)

Q24. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs. 300/- more. The sum is

(a) Rs. 5,500/-      (b) Rs. 5,000/-   (c) Rs. 5,400/-   (d) Rs. 5,300/-.

Sol. Given data :- T = 2 years, R = 3%, S.I. = 300/-,  P =?

We know the formula → S.I. = [(P×R×T)/100]

On putting the values and solving we get P = Rs. 5000/-

Option (b)

Q25. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work?

(a) 24 days            (b) 22 days         (c) 12 days         (d) 18 days.

Sol. Given data M₁ = 36 men, D₁ = 18 days, M₂ = 27 men,  D₂ = ?

Applying the formula :- M₁ × D₁ = M₂ × D₂

On putting the values and solving we get D₂ = 24 days.

Option (a)

Hope you will be able to solve these questions by your own. Visitor’s are warm heartedly welcomed with their related problems and queries. The solution of these questions (practice set) will be posted soon on this blog. More Conceptual enjoyment will be continued for you soon, just keep visiting this blog regularly and writing me your related queries through comment or forum.

Best of Luck for Exams. To be continued soon........................