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### Various types of Math Problem Answers are solved here.

1. Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck. Solution: Let the 1st paycheck be x (integer). Mrs. Rodger got a weekly raise of$ 145.

So after completing the 1st week she will get $(x+145). Similarly after completing the 2nd week she will get$ (x + 145) + $145. =$ (x + 145 + 145)

= $(x + 290) So in this way end of every week her salary will increase by$ 145.

2. The value of x + x(xx) when x = 2 is:

(a) 10, (b) 16, (c) 18, (d) 36, (e) 64

Solution:

x + x(xx)

Put the value of x = 2 in the above expression we get,

2 + 2(22)

= 2 + 2(2 × 2)

= 2 + 2(4)

= 2 + 8

= 10

3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he: (a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents Solution: Selling price of the first pipe =$1.20

Profit = 20%

Let’s try to find the cost price of the first pipe

CP = Selling price – Profit

CP = 1.20 – 20% of CP

CP = 1.20 – 0.20CP

CP + 0.20CP = 1.20

1.20CP = 1.20

CP = 1.201.201.201.20

CP = $1 Selling price of the Second pipe =$1.20

Loss = 20%

Let’s try to find the cost price of the second pipe

CP = Selling price + Loss

CP = 1.20 + 20% of CP

CP = 1.20 + 0.20CP

CP – 0.20CP = 1.20

0.80CP = 1.20

CP = 1.200.801.200.80

CP = $1.50 Therefore, total cost price of the two pipes =$1.00 + $1.50 =$2.50

And total selling price of the two pipes = $1.20 +$1.20 = $2.40 Loss =$2.50 – $2.40 =$0.10

Therefore, Mr. Jones loss 10 cents.

4. The distance light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 100 years is:

(a) 587 × 108 miles, (b) 587 × 1010 miles, (c) 587 × 10-10 miles, (d) 587 × 1012 miles, (e) 587 × 10-12 miles

Solution:

The distance of the light travels in 100 years is:

5,870,000,000,000 × 100 miles.

= 587,000,000,000,000 miles.

= 587 × 1012 miles.

5. A man has $10,000 to invest. He invests$ 4000 at 5 % and $3500 at 4 %. In order to have a yearly income of$ 500, he must invest the remainder at:

(a) 6 % , (b) 6.1 %, (c) 6.2 %, (d) 6.3 %, (e) 6.4 %

Solution:

Income from $4000 at 5 % in one year =$ 4000 of 5 %.

= $4000 × 5/100. =$ 4000 × 0.05.

= $200. Income from$ 3500 at 4 % in one year = $3500 of 4 %. =$ 3500 × 4/100.

= $3500 × 0.04. =$ 140.

Total income from 4000 at 5 % and 3500 at 4 % = $200 +$ 140 = $340. Remaining income amount in order to have a yearly income of$ 500 = $500 –$ 340.

= $160. Total invested amount =$ 4000 + $3500 =$7500.

Remaining invest amount = $10000 –$ 7500 = $2500. We know that, Interest = Principal × Rate × Time Interest =$ 160,

Principal = $2500, Rate = r [we need to find the value of r], Time = 1 year. 160 = 2500 × r × 1. 160 = 2500r 160/2500 = 2500r/2500 [divide both sides by 2500] 0.064 = r r = 0.064 Change it to a percent by moving the decimal to the right two places r = 6.4 % Therefore, he invested the remaining amount$ 2500 at 6.4 % in order to get \$ 500 income every year.

6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was:

(a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much

Solution:

Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.

We know that

Speed = Distance/Time.

Or, Time = Distance/Speed.

So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.

And times taken to covered a distance of 300 miles on his later trip = 300/3x hr.

= 100/x hr.

So we can clearly see that his new time compared with the old time was: twice as much.

7. If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is:

(a) -10.0, (b) -0.5, (c) -0.4, (d) -0.2, (e) 10.0

Solution:

(0.2)x = 2.

Taking log on both sides

log (0.2)x = log 2.

x log (0.2) = 0.3010, [since log 2 = 0.3010].

x log (2/10) = 0.3010.

x [log 2 – log 10] = 0.3010.

x [log 2 – 1] = 0.3010,[since log 10=1].

x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].

x[-0.699] = 0.3010.

x = 0.3010/-0.699.

x = -0.4306….

x = -0.4 (nearest tenth)

8. If 102y = 25, then 10-y equals:

(a) -1/5, (b) 1/625, (c) 1/50, (d) 1/25, (e) 1/5

Solution:

102y = 25

(10y)2 = 52

10y = 5

1/10y = 1/5

10-y = 1/5

9. The fraction (5x-11)/(2x2 + x – 6) was obtained by adding the two fractions A/(x + 2) and B/(2x – 3). The values of A and B must be, respectively:

(a) 5x, -11, (b) -11, 5x, (c) -1, 3, (d) 3, -1, (e) 5, -11

Solution:

10. The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:

(a) 15, (b) 20, (c) 30, (d) 32, (e) 33

Solution:

Let the three numbers be x, y and z.

Sum of the numbers is 98.

x + y + z = 98………………(i)

The ratio of the first to the second is 2/3.

x/y = 2/3.

x = 2/3 × y.

x = 2y/3.

The ratio of the second to the third is 5/8.

y/z = 5/8.

z/y = 8/5.

z = 8/5 × y.

z = 8y/5.

Put the value of x = 2y/3 and z = 8y/5 in (i).

2y/3 + y + 8y/5 = 98

49y/15 = 98.

49y = 98 × 15.

49y = 1470.

y = 1470/49.

y = 30 .

Therefore, the second number is 30.

## Maths Questions

Q.1.  300 मीटर की दूरी से क्षैffतिज तल पर स्थित किसी शीर्ष की चोटी का उन्नयन कोण 30 अंश है। शिखर की ऊंचाई ज्ञात करो –

[A] 145.20मीटर

[B] 173.20 मीटर

[C] 145मीटर

[D] 200मीटर

Q.2.  505 मीटर लंबा एक खम्भा दीवार के सहारे जमीन के 60 अंश के कोण पर टिका हुआ हो, तो खंभा दीवार की किस ऊंचाई पर पहुंचेगा –

[A] 20.36मीटर

[B] 12.99 मीटर

[C] 14मीटर

[D] 18 मीटर

Q.3. कितने वर्षो में 12 प्रतिशत वार्षिक दर से 3000 रूपए का साधारण ब्याज 1080 रूपए हो जाएगा-

[A] 9वर्ष

[B] 4वर्ष

[C] 3 वर्ष

[D] 7 वर्ष

Q.4.  किसी त्रिभुज की भुजाएं 3:4:6 के अनुपात में है, त्रिभुज है –

[A] समकोण

[B] न्यूनकोण

[C] अधिक कोण

[D] इनमें से कोई नहीं

Q.5. किसी त्रिभुज की भुजाएं 3:4:5 के अनुपात में है। त्रिभुज के सबसे बड़े कोण का माप होगा –

[A] 63 अंश

[B] 68 अंश

[C] 71अंश

[D] 75 अंश

Q.6 एक शंकु, एक गोलाई, और एक बेलन एकसमान आधार वाले हैं, और उनकी ऊंचाई भी एक समान है। तद्नुसार उ न तीनों के आयतन का अनुपात क्या होगा –

[A]  3:3:1

[B]  3:7:1

[C]  1:2:3

[D] इनमें से कोई नहीं

Q.7.  यदि किसी वस्तु के क्रय मूल्य तथा विक्रय मूल्य 10:11 के अनुपात में हो तो लाभ प्रतिशत ज्ञात कीजिए-
[A] 10
[B] 6
[C] 5
[D] 1

Q.8. प्रथम 100 धन पूर्णाकों का औसत होगा –

[A] 49.5

[B] 47

[C] 52

[D] 50.5

Q.9. यदि किसी वस्तु को 200 प्रतिशत के लाभ से बेचा जाता है, तो इसके क्रय मूल्य का इसके विक्रय मूल्य से अनुपात होगा –

[A] 1:7

[B] 2:1

[C] 1:3

[D] 6:1

Q.10.  450 रूपए में कुछ टेनिस बाल खरीदे गए। यदि प्रत्येक बाल 15 रूपए सस्ता होता  तो उतनी ही राशि में 5 बाल और खरीदे जा सकते थे। खरीदे गए बाल की संख्या ज्ञात कीजिये –
[A] 6
[B] 7

[C] 10

[D] 74

Q.11. A sum of Rs. 25000 amounts to Rs. 31000 in 4 years at the rate of simple interest what is the rate of interest?

(a) 3%

(b) 4%

(c) 5 %

(d) 6 %

(e) None of these

Q.12. Kamla took a loan of Rs. 2400 with simple interest for as many years as the rate of interest. If she paid Rs. 864 as interest at the end of the loan period, what was the rate of interest?

(a) 3.6

(b) 6

(c) 18

(d) cannot determined

(e) None of these

Q.13. What is the present worth of Rs. 264 dues in 4 years at 10% simple interest per annum?

(a) 170.20

(b) 166

(c) 188.57

(d) 175.28

Q.14. A sum fetched a total simple interest of Rs. 8016.25 at the rate of 6 p.c.p.a in 5 years what is the sum?

(a) 24720.83

(b) 26730.33

(c) 26720.83

(d) 26710.63

(e) None of these

Q.15.  8. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs 800 become in 3 years?

(a) 1020.80

(b) 1025

(c) 1052

Q. 16.  The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is

a.   74

b.   78

c.   80

d.   90

Q.17.   The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a.   3

b.   6

c.   9

d.   8.2

Q.18.  √6+√6+√6+… is equal to –

a. 2

b. 5
c. 4
d. 3

Q.19. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Q.20. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67