Top 7 Maths solutions of questions for competitive exams that you must solve

In this article, you are getting top 7 maths solutions of questions that will help you in many exams like SSC Cgl, SSC CHSL, MTS, Railways, Bank, all state exams, and many other competitive exams India. These questions will help you for quick revision in the last days of the exam.

Here, you will also get a short and simple solution that will help you for saving time in exams. Know about some important solutions of questions that can be used in your various competitive Exams, which are as follows:-

1. A spider climbed 62.5% of the height of the pole in one hour, and in the next hour, it covered 25% of the remaining height. If the pole’s height is 192 m, then the distance climbed in the second hour is -

 

 

(a) 12 m.               (b) 15 m

(c) 18 m                (d) 9 m

2. There are three inlet taps whose diameters are 1 cm, 4 cm, and 5 cm respectively. The rate of flow of the water is directly proportional to the square of the diameter. It takes 7 minutes for the smallest pipe to fill an empty tank. Find the time taken to fill an empty tank when all the three taps are opened?

(a) 14 sec.         (b) 15 sec

(c) 12 sec.         (d) 10 sec

 3. A man makes 500 articles at Rs. 6 per article. He fixes the selling price such that if only 400 articles are sold, he would have made a profit of 20% on the outlay. However, 80 articles get spoilt, and he could sell the rest articles at this price. Find his actual profit %?

 

 

(a) 36%.                 (b) 24%

(c) 26%                  (d) 33.33%

4. In an election between two candidates, 98 votes were declared invalid of total votes. The winning candidate got 62.5% of valid votes and got elected by a majority of 98 votes. Find the total votes?

(a) 392             (b) 490

(c) 588             (d) 560

5. An ore contains 25% of an alloy that has 90% iron. Other than this, in the remaining 75% of the alloy, there is 80% iron. To obtain 60kg of pure iron, the quantity of the ore needed is -

 

 

(a) 100kg                  (b) 90kg

(c) 63.63kg               (d) 72.72kg

6. Find the value of

 16/√3(Cos50°Cos10°Cos110°Cos60°)

(a) 1                 (b) 2

(c) -1                (d) -2 

7. The product of 2 numbers is 1575, and their quotient is 9/7. Then the sum of the numbers is –

 

 

(a) 74.                      (b) 96

(c) 80                       (d) 90

Solutions:-

Sol. 1. Ans. (c)

62.5% = ⅝                                                            25% = ¼                                                                    In one hr. = 192 × ⅝ = 120                                    Remaining height = 192 - 120 = 72                      The height climbed in the 2nd hr = 72 × ¼          = 18 m

Sol. 2. Ans. (d)

 

 

Rate of flow = k (diameter)²                                So the quantity filled = rate of flow × time          = k (1)² × 7 = 7k = capacity of tank                      Time taken by all taps = 7k / (1+ 16 + 25)k        = 7/42

 = 1/6 min. = 60 × ⅙ sec. = 10 sec.

Sol. 3 Ans (c)

 

 

CP = 500×6 = Rs. 3000

SP = 120% of 3000 = Rs. 3600

SP of 1 article is = 3600/400 = Rs. 9

So the actual SP = 9 × 420 = Rs. 3780

Actual profit % = (3780 – 3000)/30 = 26%

Sol. 4 Ans (b)

 

 

 

Let the total valid votes = 100

 Winning candidate got = 62.5

Loosing candidate got = (100 – 62.5) = 37.5 Majority = 62.5 – 37.5 = 25

Here it is given equal to 98

So 25 → 98

1 → 98/25

Total valid votes = 100 × 98/25 = 392

So total votes are = valid + invalid = 392 + 98 = 490

Sol. 5 Ans. (d)

 

 

 

Let the quantity of ore is 1000kg.

Here is two alloys which have quantities are respectively 250kg, 750kg

Iron in 1st alloy = 250 × 90% = 225kg

Iron in 2nd alloy = 750 × 80% = 600 kg

Total pure iron = 225 + 600 = 825kg

It is given equal to 60kg

So 825 →60

1 → 60/825

Quantity of ore is needed = 1000 × 60/825             = 72.72kg

Sol. 6 Ans. (c)

 

 

We know

[cosx° cos(60 – x)° cos (60 + x)° = ¼ Cos (3x) ]

Therefore, Cos50°Cos10°Cos110° = ¼ Cos150° = ¼ × (–√3/2) = – √3 /8

so 16 (Cos50°Cos10Cos110°Cos60°)

= 16/√3 × ( – √3/8 × 1/2 ) = –1

Sol. 7 Ans. (c)

 

 

Given xy = 1575

And x/y = 9/7

xy ÷ x/y = 1575 ÷ 9/7

y² = 1225

y = 35 and x = 45

The sum of the numbers = 45+35 = 80

I hope, guys, these questions will be beneficial for you.

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