Q1 = If the price of coal be raised by 20 % then find out by how much a householder - Maths - Probability

Question

Q1 = If the price of coal be raised by 20 % then find out by how
much a householder reduce his consumption of coal so as not to
increase his expentiture ?
Q2 A man lost 12 5 % of his money and after spending 70 % of the
remainder, has 210 left How much did the man initially have ?
Q3 A scored 25 % less thanB How much percent did B score more
than A ?
Q4P spends 75 %f his income His income is increased by 20 % and
he increased his expenditure by 10 % Findthe % age of increase in
his savings
Q5 How many Kg of pure salt be added to 30 Kg of 2 % solution of
salt and water to increase it to a 10 % solution ?
Q6 A man sells an article at a profit of 25 % If he had brought
itin 20 % less and sold it for Rs 10 50 less, he would have gained
30 % Find the cost price of the article
Q7 A man derives his income from his investment of RS 4150 at a
certain rate of simple interestandRs 3500 at 1 %higher His whole
income in 4 years is Rs 1211 Find the rate of interest ?
Q8 How much pure alcohal has to be added to 400 ml of a solution
containing15 % alcohal to change the concentration of alcohal in
the mixture of 32 % ?

Gursheen kaur. · Accepted Answer

$6) L e t C P o f a n a r t i c l e = R s x \Pr o f i t = \frac{25}{100} x = 0 25 x S P = 1 25 x n o w, w h e n h e b o u g h t i t a t 20 % l e s s, t h e n N e w C P = x - \frac{20 x}{100} = x - 0 2 x = 0 8 x a n d w h e n h e s o l d i t f o r R s 10 50 l e s s, \Rightarrow N e w S P = R s (1 25 x - 10 50) t h e r e f o r e, a c c t o q u e s t i o n, \frac{N e w S P - N e w C P}{N e w C P} = \frac{30}{100} \Rightarrow \frac{1 25 x - 10 50 - 0 8 x}{0 8 x} = \frac{3}{10} \Rightarrow \frac{0 45 x - 10 50}{0 8 x} = \frac{3}{10} \Rightarrow 10 (0 45 x - 10 5) = 3 * 0 8 x \Rightarrow 4 5 x - 105 = 2 4 x \Rightarrow 4 5 x - 2 4 x = 105 \Rightarrow 2 1 x = 105 \Rightarrow x = \frac{105}{2 1} = 50 \Rightarrow x = R s 50 = C P o f a n a r t i c l e$ $7) G i v e n : I n c o m e f o r 4 y e a r s = R s 1211 t h e r e f o r e, I n c o m e f o r 1 y e a r = \frac{R s 1211}{4} A l s o, i t i s g i v e n t h a t, r a t e o f int e r e s t f o r R s 3500 i s 1 % h i g h e r, i t m e a n s i f 1 % o n R s 3500 i s \min u s f r o m \frac{R s 1211}{4}, t h e n t h e r e m a i n d e r i s 1 y e a r int e r e s t o n (R s 4150 + R s 3500) a t l o w e r r a t e o f int e r e s t t h e r e f o r e, I n t e r e s t o n (R s 4150 + R s 3500 = R s 7650) = R s \frac{1211}{4} - 1 % o n R s 3500 I n t e r e s t o n (R s 4150 + R s 3500 = R s 7650) = R s (\frac{1211}{4} - 35) = R s (\frac{1211 - 140}{4}) = R s \frac{1071}{4} I n t e r e s t o n R s 100 = \frac{1071}{4} * \frac{100}{7650} = \frac{7}{2} = R s 3 \frac{1}{2} t h u s, l o w e r r a t e o f int e r e s t = 3 \frac{1}{2} % & h i g h e r r a t e o f f int e r e s t = (3 \frac{1}{2} + 1) % = 4 \frac{1}{2} %$ $8) G i v e n : S o l u t i o n c o n t a i n s 15 % a l c o h o l, t h e n v o l u m e = 400 m l i t m e a n s, q t y o f a l c o h o l i n 400 m l s o l u t i o n = 400 * \frac{15}{100} = 60 m l o f a l c o h o l a n d q t y o f w a t e r = 400 - 60 = 340 m l N o w, l e t t h e x m l b e t h e a l c o h o l t o b e a d d e d t o m a k e s t r e n g t h 32 % t h e r e f o r e, a c c t o t h e q u e s t i o n, \Rightarrow \frac{60 + x}{400 + x} = \frac{32}{100} \Rightarrow 100 (60 + x) = 32 (400 + x) \Rightarrow 6000 + 100 x = 12800 + 32 x \Rightarrow 100 x - 32 x = 12800 - 6000 \Rightarrow 68 x = 6800 \Rightarrow x = \frac{6800}{68} \Rightarrow x = 100 t h u s, 100 m l b e t h e a l c o h o l t o b e a d d e d t o m a k e s t r e n g t h 32 %$