बट्ट10% व 15%के समतुल्य बट्टा ज्ञात कीजिए।
एक दुकानदार एक मेज के अंकित मूल्य पर 10% की छूट देता है और 25% का लाभ कमाता है। यदि मेज का अंकित मूल्य
1250₹ हो, तो मेज का क्रय मूल्य क्या होगा?
मेएल
Answers
Answer:
Answer:
Solution:
Let the number of Rs.50 notes be A and the number of Rs.100 notes be B
According to the given information,
A + B = 25 ……………………………………………………………………….. (i)
50A + 100B = 2000 ………………………………………………………………(ii)
When equation (i) is multiplied with (ii) we get,
50A + 50B = 1250 …………………………………………………………………..(iii)
Subtracting the equation (iii) from the equation (ii) we get,
50B = 750
B = 15
Substituting in the equation (i) we get,
A = 10
Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100.Answer:
Solution:
Let the fixed charge be Rs x and per km charge be Rs y.
According to the question,
x + 10y = 105 …………….. (1)
x + 15y = 155 …………….. (2)
From (1), we get x = 105 – 10y ………………. (3)
Substituting the value of x in (2), we get
105 – 10y + 15y = 155
5y = 50
y = 10 …………….. (4)
Putting the value of y in (3), we get
x = 105 – 10 × 10 = 5
Hence, fixed charge is Rs 5 and per km charge = Rs 10
Charge for 25 km = x + 25y = 5 + 250 = Rs 255
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Answer:
Solution :-
↦Let, the first number be x
↦ Second number be x + 1
↦ Third number be x + 2
↦ Fourth number be x + 3
↦ Fifth number be x + 4
↦ And, the addition of five consecutive numbers is 270.
According to the question,
⇒ x + x + 1 + x + 2 + x + 3 + x + 4 = 270
⇒ x + x + x + x + x + 1 + 2 + 3 + 4 = 270
⇒ 5x + 10 = 270
⇒ 5x = 270 - 10
⇒ 5x = 260
⇒ < /p > < p > x = < /p > < p > \dfrac{260}{5} \: ⇒</p><p>x=</p><p>
5
260
➠x=52
Hence, the required numbers are,
✦ First number = x = 52
✦ Second number = x + 1 = 52 + 1 = 53
✦ Third number = x + 2 = 52 + 2 = 54
✦ Fourth number = x + 3 = 52 + 3 = 55
✦ Fifth number = x + 4 = 52 + 4 = 56
But, in question we have to find the addition of second and fifth number,
➳ Second number + Fifth number
➳ 53 + 56
\implies < /p > < p > 109 \: ⟹</p><p>109
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hopethishelpsu
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Hope this helps u